Solve Using the Quadratic Formula (x-3)(x+4)=0

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

Step 1

Simplify the left side.

Simplify $(x - 3) (x + 4)$ .

Expand $(x - 3) (x + 4)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

$x \cdot x + x \cdot 4 - 3 (x + 4) = 0$

Apply the distributive property.

$x \cdot x + x \cdot 4 - 3 x - 3 \cdot 4 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 4 - 3 x - 3 \cdot 4 = 0$

Move $4$ to the left of $x$ .

$x^{2} + 4 \cdot x - 3 x - 3 \cdot 4 = 0$

Multiply $- 3$ by $4$ .

$x^{2} + 4 x - 3 x - 12 = 0$

Subtract $3 x$ from $4 x$ .

$x^{2} + x - 12 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = 1$ , and $c = - 12$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot - 12)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 12}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 12$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 12}}{2 \cdot 1}$

Multiply $- 4$ by $- 12$ .

$x = \frac{- 1 \pm \sqrt{1 + 48}}{2 \cdot 1}$

Add $1$ and $48$ .

$x = \frac{- 1 \pm \sqrt{49}}{2 \cdot 1}$

Rewrite $49$ as $7^{2}$ .

$x = \frac{- 1 \pm \sqrt{7^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 1 \pm 7}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm 7}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 3, - 4$

Solve Using the Quadratic Formula (x-3)(x+4)=0

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