How to Solve Quadratic Equations: 4 Methods Explained
A quadratic equation is any equation that can be written in the form:
Solve Quadratic Equations: 4 Methods Explained: what to notice and how to work it
What to notice first
Common student mistake
Key formulas and cues
A reliable path
- Read the formFactored, standard, and vertex forms reveal different features.
- Choose the methodFactor when friendly, complete the square for structure, or use the formula when needed.
- Connect to the graphRoots are x-intercepts and the vertex is the minimum or maximum point.
Worked examples
Factor and solve
- Factor into (x – 3)(x – 4).
- Set each factor equal to zero.
- Solve both small equations.
Find the axis
- Use x = -b/(2a).
- Here a = 2 and b = -8.
- Compute 8/4.
Try one before moving on
Solve Quadratic Equations: 4 Methods Explained: pop-up practice
\[ax^2 + bx + c = 0\]
where \(a \neq 0\). Quadratics show up everywhere: projectile motion, area problems, profit maximization, and almost every Algebra I final exam.
There are four reliable methods to solve them. This guide walks through each one — when to use it, how to use it, and where students trip up.
Quadratic Equations at a Glance
\[ax^2 + bx + c = 0\]
Solving a quadratic means finding the roots: the x-values where the parabola crosses the x-axis. Those roots are also called zeros or solutions.
First put the equation in standard form, identify \(a\), \(b\), and \(c\), then choose the fastest method.
Which Method Should You Use?
Try It: Quadratic Solver and Graph Check
Enter \(a\), \(b\), and \(c\) for \(ax^2+bx+c=0\). The tool shows the discriminant, the roots, and a quick graph so you can connect the algebra to the picture.
Hidden Quadratics: Put It in Standard Form First
Many test questions do not start as \(ax^2+bx+c=0\). Move everything to one side before choosing a method.
| Original equation | Move to standard form | \(a\), \(b\), \(c\) |
|---|---|---|
| \(x^2 = 3x – 1\) | \(x^2 – 3x + 1 = 0\) | \(a=1\), \(b=-3\), \(c=1\) |
| \(2(x^2 – 2x)=5\) | \(2x^2 – 4x – 5 = 0\) | \(a=2\), \(b=-4\), \(c=-5\) |
| \(x(x-1)=3\) | \(x^2 – x – 3 = 0\) | \(a=1\), \(b=-1\), \(c=-3\) |
Quick Check: Try One Before You Scroll
Solve \(x^2 – 7x + 12 = 0\)
Factor: \((x-3)(x-4)=0\), so \(x=3\) or \(x=4\).
Solve \(x^2 = 81\)
Use square roots: \(x=\pm 9\).
How many real solutions does \(x^2+2x+5=0\) have?
The discriminant is \(2^2-4(1)(5)=-16\), so there are no real solutions.
The 4 Methods (and When to Use Each)
| Method | Best when... |
|---|---|
| Factoring | The quadratic factors easily over integers. |
| Square root property | The equation has no \(bx\) term (form \(ax^2 + c = 0\)). |
| Completing the square | You want to convert to vertex form, or the leading coefficient is 1 with even \(b\). |
| Quadratic formula | Always works. Use when factoring fails. |
Train yourself to ask: "Can this factor? If not, can I use the square root? If not, use the formula."
Method 1: Factoring
Steps: 1. Set the equation to \(= 0\). 2. Factor the polynomial. 3. Set each factor equal to 0. 4. Solve each.
Example 1
Solve \(x^2 - 5x + 6 = 0\).
Factor: \((x - 2)(x - 3) = 0\). So \(x - 2 = 0\) or \(x - 3 = 0\). \(x = 2\) or \(x = 3\).
Example 2
Solve \(2x^2 + 5x - 3 = 0\).
Factor: \((2x - 1)(x + 3) = 0\). \(x = \dfrac{1}{2}\) or \(x = -3\).
Example 3 (with GCF)
Solve \(3x^2 - 12x = 0\).
Factor: \(3x(x - 4) = 0\). \(x = 0\) or \(x = 4\).
The zero product property is the key: if a product equals zero, at least one of the factors must be zero.
Method 2: Square Root Property
Use when the equation looks like \(x^2 = k\) or \((x - h)^2 = k\) — no \(bx\) term.
Rule: \(x^2 = k \implies x = \pm\sqrt{k}\).
Example 4
Solve \(x^2 = 49\). \(x = \pm 7\).
Example 5
Solve \((x - 3)^2 = 16\). \(x - 3 = \pm 4\). \(x = 3 + 4 = 7\) or \(x = 3 - 4 = -1\).
Example 6
Solve \(2x^2 - 50 = 0\). \(2x^2 = 50\). \(x^2 = 25\). \(x = \pm 5\).
Don't forget the \(\pm\)! That is the most common mistake.
Recommended Practice Resources
Method 3: Completing the Square
Use when you want vertex form, or when factoring is awkward.
Steps for \(x^2 + bx + c = 0\): 1. Move \(c\) to the right side. 2. Take half of \(b\), square it, add it to both sides. 3. Write the left side as a perfect square. 4. Square root both sides (with \(\pm\)). 5. Solve for \(x\).
Example 7
Solve \(x^2 + 6x - 7 = 0\).
Step 1: \(x^2 + 6x = 7\). Step 2: half of 6 is 3; squared is 9. Add 9 to both sides: \(x^2 + 6x + 9 = 16\). Step 3: \((x + 3)^2 = 16\). Step 4: \(x + 3 = \pm 4\). Step 5: \(x = -3 + 4 = 1\) or \(x = -3 - 4 = -7\).
Example 8 (leading coefficient ≠ 1)
Solve \(2x^2 + 8x - 10 = 0\).
Divide both sides by 2: \(x^2 + 4x - 5 = 0\). Move: \(x^2 + 4x = 5\). Add \((\frac{4}{2})^2 = 4\): \(x^2 + 4x + 4 = 9\). Square: \((x + 2)^2 = 9\). Root: \(x + 2 = \pm 3\). \(x = 1\) or \(x = -5\).
Completing the square is also how you derive the quadratic formula and how you convert standard form to vertex form for graphing.
Method 4: The Quadratic Formula
The universal solver. Works on every quadratic equation, every time.
\[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Steps: 1. Put the equation in \(ax^2 + bx + c = 0\) form. 2. Identify \(a\), \(b\), \(c\). 3. Plug into the formula. 4. Simplify.
Example 9
Solve \(x^2 + 5x + 6 = 0\) using the formula.
\(a = 1\), \(b = 5\), \(c = 6\).
\[x = \dfrac{-5 \pm \sqrt{25 - 24}}{2} = \dfrac{-5 \pm 1}{2}\]
\(x = -2\) or \(x = -3\). (Same as factoring would give.)
Example 10
Solve \(2x^2 - 3x - 1 = 0\).
\(a = 2\), \(b = -3\), \(c = -1\).
\[x = \dfrac{3 \pm \sqrt{9 + 8}}{4} = \dfrac{3 \pm \sqrt{17}}{4}\]
The answers are irrational — exactly the case where you must use the formula.
Example 11 (no real solutions)
Solve \(x^2 + 2x + 5 = 0\).
\(a = 1\), \(b = 2\), \(c = 5\).
\[x = \dfrac{-2 \pm \sqrt{4 - 20}}{2} = \dfrac{-2 \pm \sqrt{-16}}{2}\]
\(\sqrt{-16}\) is not a real number. No real solutions. (In complex form: \(x = -1 \pm 2i\).)
The Discriminant
Inside the square root sits the discriminant:
\[D = b^2 - 4ac\]
It tells you how many real solutions the equation has:
- $D > 0$: two real solutions.
- \(D = 0\): one real solution (a repeated root).
- $D < 0$: no real solutions (two complex solutions).
Memorize this. It comes up directly on many tests.
Common Mistakes
Forgetting the \(\pm\)
The square root property always produces two answers (unless the value is 0).
Sign errors with the formula
\(-b\) is positive when \(b\) is negative. If \(b = -3\), then \(-b = 3\). Watch carefully.
Forgetting to set \(= 0\)
You cannot factor or use the formula unless the equation equals zero. Move everything to one side first.
Mishandling fractions
\(x = \dfrac{4 \pm 2}{2}\) is \(x = 3\) or \(x = 1\) — not \(x = 6\) or \(x = 2\). Divide the whole numerator.
Half of an odd number
If \(b = 5\), half of \(b\) is $2.5$, and \((2.5)^2 = 6.25\). Use fractions: half is \(\dfrac{5}{2}\), squared is \(\dfrac{25}{4}\).
Forgetting to simplify the radical
\(\sqrt{12} = 2\sqrt{3}\), not just \(\sqrt{12}\).
Real-World Example: Projectile Motion
A ball is thrown upward from a 6-foot platform at 32 ft/s. Its height after \(t\) seconds is:
\(h(t) = -16t^2 + 32t + 6\)
When does the ball hit the ground? Set \(h(t) = 0\):
\(-16t^2 + 32t + 6 = 0\)
Using the quadratic formula with \(a = -16\), \(b = 32\), \(c = 6\):
\[t = \dfrac{-32 \pm \sqrt{1024 + 384}}{-32} = \dfrac{-32 \pm \sqrt{1408}}{-32}\]
\(\sqrt{1408} \approx 37.52\).
\(t \approx \dfrac{-32 - 37.52}{-32} \approx 2.17\) seconds. (Reject the negative time.)
The ball hits the ground after about 2.17 seconds.
Free Resources and Practice Path
Use these next so the idea turns into speed and confidence:
5-Day Quadratics Practice Path
- Day 1: Factor simple quadratics and use the zero-product property.
- Day 2: Solve square-root property problems such as \(x^2=k\) and \((x-h)^2=k\).
- Day 3: Complete the square and connect the result to vertex form.
- Day 4: Use the quadratic formula and classify answers with the discriminant.
- Day 5: Mix all methods and time yourself: aim for under two minutes per problem.
Frequently Asked Questions
Which method should I learn first? Factoring. It's fastest when it works, and it builds your intuition for the others.
Do I always have to memorize the quadratic formula? Yes. It's the universal backup and it appears on every standardized test.
What's the difference between a quadratic equation and a quadratic function? A function is \(y = ax^2 + bx + c\) (a graph). An equation is \(ax^2 + bx + c = 0\) (find the \(x\)-intercepts).
Can a quadratic have one solution? Yes — when the discriminant equals 0. The graph touches the \(x\)-axis at exactly one point.
Are imaginary roots important? For Algebra II and beyond, yes. For Algebra I and most standardized tests, "no real solutions" is enough.
How do I know which method to use on a test? Try factoring first (10 seconds to check). If it doesn't factor cleanly, jump to the quadratic formula.
You Have Every Tool You Need
Quadratics reward practice. Drill each method until each takes under two minutes. Then mix them. By exam day, you will look at any quadratic, choose the right method instantly, and solve it without breaking a sweat.
Keep Practicing With the Right Resources
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