Quadratic Equations — Factoring, Formula, and Examples

By MathPal Editorial Team·Published Apr 17, 2026

Grade: 9 | Topic: Algebra

What You Will Learn

A quadratic equation is any equation that can be written in the form $ax^{2} + bx + c = 0$. You will encounter quadratics in physics, geometry, finance, and many other fields. This guide teaches you two main methods for solving them — factoring and the quadratic formula — and introduces the discriminant, which tells you how many solutions to expect before you solve.

Theory

Standard form

A quadratic equation has the form:

$ax^{2} + bx + c = 0$

where $a$, $b$, and $c$ are real numbers and $a \neq 0$. The "2" in $x^{2}$ is what makes it quadratic (from the Latin quadratus, meaning "square").

Examples in standard form:

• $x^{2} - 5x + 6 = 0$ (here $a = 1$, $b = -5$, $c = 6$)
• $2x^{2} + 3x - 2 = 0$ (here $a = 2$, $b = 3$, $c = -2$)

If the equation is not in standard form, rearrange it first by moving all terms to one side.

Method 1 — Solving by factoring

Factoring works when you can rewrite $ax^{2} + bx + c$ as a product of two binomials. The idea relies on the zero-product property: if $pq = 0$, then $p = 0$ or $q = 0$.

Steps:

1. Write the equation in standard form.
2. Factor the left side into two binomials.
3. Set each binomial equal to zero.
4. Solve each resulting linear equation.

For the simple case where $a = 1$, you need two numbers that multiply to $c$ and add to $b$.

Method 2 — The quadratic formula

When factoring is difficult, the quadratic formula always works:

$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

The $\pm$ symbol means you compute two values — one with $+$ and one with $-$ — giving you (up to) two solutions.

The discriminant

The expression under the square root, $b^{2} - 4ac$, is called the discriminant (often written $D$ or $\Delta$):

$D = b^{2} - 4ac$

• $D > 0$: two distinct real solutions.
• $D = 0$: exactly one real solution (a repeated root).
• $D < 0$: no real solutions (the parabola does not cross the $x$-axis).

Checking the discriminant first can save time — you will know what kind of answer to expect.

Worked Examples

Example 1 — Factoring a simple quadratic

Solve $x^{2} - 5x + 6 = 0$.

Step 1: Find two numbers that multiply to $6$ and add to $-5$. Those numbers are $-2$ and $-3$.

Step 2: Factor: $(x - 2)(x - 3) = 0$.

Step 3: Set each factor to zero:

• $x - 2 = 0 \Rightarrow x = 2$
• $x - 3 = 0 \Rightarrow x = 3$

Answer: $x = 2$ or $x = 3$.

Example 2 — Factoring with a leading coefficient

Solve $2x^{2} + 5x - 3 = 0$.

Step 1: Multiply $a \times c = 2 \times (-3) = -6$. Find two numbers that multiply to $-6$ and add to $5$: those are $6$ and $-1$.

Step 2: Rewrite the middle term: $2x^{2} + 6x - x - 3 = 0$.

Step 3: Factor by grouping:

$2x(x + 3) - 1(x + 3) = 0$

$(2x - 1)(x + 3) = 0$

Step 4: Solve:

• $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$
• $x + 3 = 0 \Rightarrow x = -3$

Answer: $x = \frac{1}{2}$ or $x = -3$.

Example 3 — Using the quadratic formula

Solve $3x^{2} - 2x - 4 = 0$.

Step 1: Identify $a = 3$, $b = -2$, $c = -4$.

Step 2: Compute the discriminant:

$D = (-2)^{2} - 4(3)(-4) = 4 + 48 = 52$

Since $D > 0$, there are two real solutions.

Step 3: Apply the formula:

$x = \frac{-(-2) \pm \sqrt{52}}{2(3)} = \frac{2 \pm \sqrt{52}}{6}$

Simplify $\sqrt{52} = 2\sqrt{13}$:

$x = \frac{2 \pm 2\sqrt{13}}{6} = \frac{1 \pm \sqrt{13}}{3}$

Answer: $x = \frac{1 + \sqrt{13}}{3} \approx 1.54$ or $x = \frac{1 - \sqrt{13}}{3} \approx -0.87$.

Example 4 — Using the discriminant to predict solutions

How many real solutions does $x^{2} + 4x + 4 = 0$ have?

Step 1: $D = 4^{2} - 4(1)(4) = 16 - 16 = 0$.

Since $D = 0$, there is exactly one repeated root.

Step 2: Solve: $x = \frac{-4}{2} = -2$.

Answer: $x = -2$ (a double root). The equation factors as $(x + 2)^{2} = 0$.

Common Mistakes

Mistake 1 — Forgetting to set the equation to zero

❌ Solve $x^{2} = 3x + 10$ by factoring $x^{2}$ directly.

✅ First rearrange: $x^{2} - 3x - 10 = 0$, then factor $(x - 5)(x + 2) = 0$.

Mistake 2 — Sign errors in the quadratic formula

❌ For $x^{2} - 6x + 5 = 0$, writing $x = \frac{6 \pm \sqrt{36 - 20}}{2}$ but computing $b^{2}$ as $6^{2}$ instead of $(-6)^{2}$.

✅ Both give $36$ in this case, but always substitute the sign of $b$ into $-b$. Here $-b = -(-6) = 6$.

Mistake 3 — Dividing by $2a$ incorrectly

❌ For $2x^{2} + 4x - 6 = 0$: $x = \frac{-4 \pm \sqrt{64}}{2}$ (dividing by $2$ instead of $2a = 4$).

✅ $x = \frac{-4 \pm 8}{4}$, giving $x = 1$ or $x = -3$.

Practice Problems

Problem 1: Solve $x^{2} + 7x + 12 = 0$ by factoring.

Show Answer

$(x + 3)(x + 4) = 0$

$x = -3$ or $x = -4$

Problem 2: Solve $x^{2} - 9 = 0$.

Show Answer

$(x - 3)(x + 3) = 0$

$x = 3$ or $x = -3$

Problem 3: Use the quadratic formula to solve $x^{2} + 2x - 5 = 0$.

Show Answer

$a = 1$, $b = 2$, $c = -5$

$D = 4 + 20 = 24$

$x = \frac{-2 \pm \sqrt{24}}{2} = \frac{-2 \pm 2\sqrt{6}}{2} = -1 \pm \sqrt{6}$

$x \approx 1.45$ or $x \approx -3.45$

Problem 4: Find the discriminant of $2x^{2} - 4x + 2 = 0$ and state how many real solutions it has.

Show Answer

$D = (-4)^{2} - 4(2)(2) = 16 - 16 = 0$

There is exactly one real solution (a double root): $x = 1$.

Problem 5: Solve $3x^{2} + x - 2 = 0$ by any method.

Show Answer

Factor: $(3x - 2)(x + 1) = 0$

$x = \frac{2}{3}$ or $x = -1$

Summary

• A quadratic equation has the form $ax^{2} + bx + c = 0$ with $a \neq 0$.
• Factoring uses the zero-product property — rewrite the quadratic as two binomials and set each to zero.
• The quadratic formula $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ works for all quadratics.
• The discriminant $D = b^{2} - 4ac$ tells you whether there are 2, 1, or 0 real solutions.
• Always write the equation in standard form before solving.

Related Topics

• Linear Equations — simpler one-step and two-step equations
• Exponents and Powers — review exponent rules used in quadratics
• Square Roots and Cube Roots — understand the square root in the formula

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