How to Solve Quadratic Equations: 4 Methods Explained

A quadratic equation is any equation that can be written in the form:

\[ax^2 + bx + c = 0\]

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.

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The 4 Methods (and When to Use Each)

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.

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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 \(2x^2 – 3x – 1 = 0\).

\(a = 2\), \(b = -3\), \(c = -1\).

\[x = \dfrac{3 \pm \sqrt{9 + 8}}{4} = \dfrac{3 \pm \sqrt{17}}{4}\]

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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

Effortless Math has a full quadratic library:

• Algebra 1 Worksheets — quadratic problems by method, with answers.
• Math Topics Library — quadratic equations explained.
• Algebra 1 eBooks — complete Algebra workbooks.

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.

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