Quadratic Equation Solver (Free Step-by-Step Tool)
AlgebraFreeNo sign-up
Use the quadratic equation solver to solve ax² + bx + c = 0 with every step shown free. For 2x² + 5x − 3 = 0, a = 2, b = 5 and c = −3, so the discriminant is 5² − 4(2)(−3) = 49. The quadratic formula gives x = (−5 ± 7)/4, so x = 1/2 or x = −3.
Show steps: Factoring (6)
- Write the equation in standard form with a = 2, b = 5, c = −3: 2x2 + 5x − 3 = 0.
- Use the AC method (factoring by grouping). Multiply a · c = 2 · (−3) = −6. Find two numbers that multiply to −6 and add to b = 5: 6 and −1.
- Split the middle term 5x into 6x and −x: 2x2 + 6x − x − 3 = 0.
- Factor each pair: 2x(x + 3) − 1(x + 3) = 0.
- Factor out the common binomial (x + 3): (2x − 1)(x + 3) = 0.
- Zero product property: 2x − 1 = 0 gives x = 1/2, and x + 3 = 0 gives x = −3.
Vertex, axis and intercepts (6)
- The graph of y = 2x2 + 5x − 3 is a parabola. Its axis of symmetry is the vertical line x = −b/(2a) = −5/(2·2) = −5/4.
- Find the vertex by substituting x = −5/4: y = 2(−5/4)2 + 5(−5/4) − 3 = 25/8 − 25/4 − 3 = −49/8. The vertex is (−5/4, −49/8).
- Vertex form y = a(x − h)2 + k with h = −5/4 and k = −49/8: y = 2(x + 5/4)2 − 49/8.
- Because a = 2 is positive, the parabola opens upward and the vertex is its lowest point: the minimum value is y = −49/8, and the range is y ≥ −49/8.
- y-intercept: set x = 0, so y = c = −3. The parabola crosses the y-axis at (0, −3).
- x-intercepts: the real solutions, (−3, 0) and (1/2, 0). They are the same distance from the axis of symmetry x = −5/4.
Working on more than quadratics? The free Math Solver solves equations, inequalities, systems, trig and calculus problems with free step-by-step work.
How to use the quadratic equation solver
- Enter a, b and c from the standard form ax² + bx + c = 0 (a cannot be 0), or open the Type the equation tab and type it as you see it, such as 3x^2 = 2x + 5 or x(x + 4) = 12. The solver rearranges it for you.
- Pick the method you want to see in the steps: factoring, the quadratic formula or completing the square. Auto uses the simplest one that works.
- Press Solve, or just keep typing; the answer updates as you go.
- Read the yellow check box, open Show steps, and use Copy result or Link to save or share the problem. A link such as ?a=2&b=5&c=-3 reopens the same equation.
Coefficients can be whole numbers, negatives, fractions such as 1/2 or decimals such as 0.75. All arithmetic is exact, so answers stay as fractions and simplified radicals, with decimals beside them.
What the solver shows you
- Roots: exact rational, irrational (simplified radical) or complex roots written with i.
- Discriminant b² − 4ac and what it tells you about the roots.
- Factored form when the quadratic factors over the integers.
- Vertex, axis of symmetry and vertex form y = a(x − h)² + k, plus whether the parabola opens up or down.
- Intercepts: the y-intercept (0, c) and the x-intercepts, if any.
- Graph of the parabola with the roots, vertex, y-intercept and axis of symmetry marked.
- Check by substitution: each root is put back into the equation, including irrational and complex roots, and the result is shown to equal 0.
The quadratic formula and the discriminant
Every quadratic equation ax² + bx + c = 0 with a ≠ 0 can be solved with the quadratic formula:
x = (−b ± √(b² − 4ac)) / (2a)
The expression under the square root, D = b² − 4ac, is the discriminant. You can read the type of answer from it before you finish the problem.
| Discriminant | Roots | Graph | Example |
|---|---|---|---|
| Positive perfect square | Two rational roots | Crosses the x-axis twice | x² − 2x − 8 = 0: D = 36, x = 4 or −2 |
| Positive, not a perfect square | Two irrational roots | Crosses the x-axis twice | x² − 2x − 1 = 0: D = 8, x = 1 ± √2 |
| Zero | One double root | Touches the x-axis at the vertex | x² − 2x + 1 = 0: D = 0, x = 1 |
| Negative | Two complex roots | Never meets the x-axis | x² − 2x + 5 = 0: D = −16, x = 1 ± 2i |
With integer coefficients, a perfect-square discriminant is also the test for factoring: the quadratic factors over the integers exactly when b² − 4ac is a perfect square.
Three ways to solve a quadratic equation
Factoring
Factoring is the quickest method when it works. For x² − 5x + 6 = 0, find two numbers that multiply to 6 and add to −5: −2 and −3. So (x − 2)(x − 3) = 0, and the zero product property gives x = 2 or x = 3. When a is not 1, the solver uses the AC method: for 2x² + 5x − 3, multiply a · c = −6, split 5x into 6x − x, and group: 2x(x + 3) − 1(x + 3) = (2x − 1)(x + 3).
The quadratic formula
The formula always works. For 2x² + 3x − 4 = 0, D = 3² − 4(2)(−4) = 9 + 32 = 41, which is not a perfect square, so x = (−3 ± √41)/4. That is about 0.8508 and −2.3508.
Completing the square
Completing the square turns the left side into a perfect square. For x² + 6x + 4 = 0, move the constant: x² + 6x = −4. Half of 6 is 3, and 3² = 9, so add 9 to both sides: x² + 6x + 9 = 5, which is (x + 3)² = 5. Take square roots: x + 3 = ±√5, so x = −3 ± √5, about −0.7639 and −5.2361. This method is also how the quadratic formula is derived and how vertex form is found.
Worked examples
Two real roots: 2x² + 5x − 3 = 0
D = 25 + 24 = 49 and √49 = 7, so x = (−5 + 7)/4 = 1/2 and x = (−5 − 7)/4 = −3. Check x = −3: 2(9) − 15 − 3 = 0.
One repeated root: x² − 4x + 4 = 0
D = 16 − 16 = 0, so there is one double root, x = 4/2 = 2. The equation is (x − 2)² = 0, and the parabola touches the x-axis at its vertex (2, 0).
Complex roots: x² + x + 1 = 0
D = 1 − 4 = −3. A negative discriminant means no real solutions; with √(−3) = i√3, x = (−1 ± i√3)/2, which is −0.5 ± (√3/2)i, or about −0.5 ± 0.866i.
An equation that needs rearranging: x(x + 4) = 12
Distribute and move 12 to the left: x² + 4x − 12 = 0. The numbers 6 and −2 multiply to −12 and add to 4, so (x + 6)(x − 2) = 0 and x = −6 or x = 2.
Vertex form, axis of symmetry and the graph
The graph of y = ax² + bx + c is a parabola. Its axis of symmetry is x = −b/(2a), and the vertex lies on that line. For y = 2x² + 5x − 3, the axis is x = −5/4. Substituting gives y = 2(25/16) − 25/4 − 3 = −49/8, so the vertex is (−5/4, −49/8) and the vertex form is y = 2(x + 5/4)² − 49/8. Because a is positive, the parabola opens upward and −49/8 is its minimum value. The y-intercept is (0, c) = (0, −3), and the x-intercepts are the real roots, (−3, 0) and (1/2, 0), which sit the same distance from the axis.
Common mistakes to avoid
| Mistake | How to catch it |
|---|---|
| Reading b and c before the equation equals 0 | Move every term to one side first; in 3x² = 2x + 5, b is −2 and c is −5. |
| Dropping a negative sign | Write b in parentheses: (−5)² = 25, not −25. |
| Dividing only part of the top by 2a | The whole numerator −b ± √D sits over 2a. |
| Rounding too early | Keep √41 exact until the last step, then round. |
| Saying “no solution” when D < 0 | There are no real solutions, but there are two complex ones. |
| Forgetting the ± sign | x² = 16 has two answers, 4 and −4. |
Using the quadratic equation solver for homework
Copy the problem exactly, with parentheses where the original has them, and estimate first: look at the sign of the discriminant to know whether to expect real or complex roots. Solve by hand, then compare each line with the steps above and find the first place they differ. When a test allows no calculator, practise the method your teacher expects; switching the method menu shows the same problem solved three ways.
Quadratic equations practice quiz: 10 questions
Test yourself with 10 multiple-choice questions. Choose an answer and press Submit to see if you are right and why. Your score appears at the end.
Find two numbers that multiply to 10 and add to −7: −2 and −5. So (x − 2)(x − 5) = 0, giving x = 2 or x = 5.
b² − 4ac = (−4)² − 4(3)(2) = 16 − 24 = −8. A common slip is squaring −4 as −16, which gives −40; (−4)² is 16.
With D = 0 the ± part of the formula is ± 0, so both roots are the same number. The parabola touches the x-axis at its vertex.
x² + 6x = −4. Add (6/2)² = 9 to both sides: (x + 3)² = 5. So x + 3 = ±√5 and x = −3 ± √5.
D = 9 + 16 = 25, so x = (−3 ± 5)/4, which gives 2/4 = 1/2 and −8/4 = −2.
D = 4 − 40 = −36 and √(−36) = 6i, so x = (−2 ± 6i)/2 = −1 ± 3i. There are no real solutions, but there are two complex ones.
The axis is x = −b/(2a) = 6/2 = 3. Then y = 9 − 18 + 5 = −4, so the vertex is (3, −4).
Take the square root of both sides, remembering both signs: x = ±√18 = ±√(9 · 2) = ±3√2.
Subtract 2x and 8 from both sides: 3x² − 2x − 8 = 0. Reading b and c before moving the terms keeps the wrong signs.
Quiz complete
Frequently asked questions
What is the solution of a quadratic equation?
The solutions, or roots, are the values of x that make ax² + bx + c equal 0. Real roots are where the parabola crosses the x-axis. For x² − 9 = 0 the roots are 3 and −3.
How do you solve a quadratic equation for x?
Write it in standard form ax² + bx + c = 0, then factor, complete the square or use the quadratic formula. For x² + 4x − 12 = 0, factoring gives (x + 6)(x − 2) = 0, so x = −6 or x = 2.
What is the quadratic formula?
x = (−b ± √(b² − 4ac))/(2a). The discriminant b² − 4ac tells you whether there are two real roots, one double root or two complex roots.
How do you solve a quadratic equation by factoring?
Write it as a product of two binomials and set each equal to 0. x² − 5x + 6 = (x − 2)(x − 3), so x = 2 or x = 3. If the discriminant is not a perfect square, it does not factor over the integers; use the formula.
How do you solve a quadratic equation on a calculator?
Enter a, b and c in the solver above, or type the whole equation. On a handheld calculator, find b² − 4ac first, take its square root, then work out (−b + √D)/(2a) and (−b − √D)/(2a), using parentheses around the numerator.
How can you tell if an equation is quadratic?
After expanding and collecting terms, the highest power of the variable is 2. Some are hidden: x(x + 4) = 12 and (x − 3)² = 16 are quadratic, while x² = x² + 2x is linear because the x² terms cancel.
Related calculators
- Algebra calculator
- Completing the square calculator
- Factoring calculator
- Parabola calculator
- Graphing calculator
- Quadratic inequality calculator
- Quadratic Formula Calculator
More algebra calculators
Related to This Article
More math articles
- Area Models Unveiled: How to Complete Decimal Division Equations
- Cleaning, Disinfection, and Sterilization
- Flashback and Foreshadowing
- Life Course, New Roles, and Resocialization
- The Best Grade 7 Math Book for Tennessee Students
- Is the PERT Test Hard?
- A Comprehensive Collection of Free PERT Math Practice Tests
- Free Grade 7 English Worksheets for Idaho Students
- Functionalist, Conflict, Feminist, and Interactionist Views of Family
- Education as a Social Institution
What people say about "Quadratic Equation Solver: Free Step-by-Step Tool"?
No one replied yet.