Solve Using the Quadratic Formula (x+5)(x-3)=0

$(x + 5) (x - 3) = 0$

Step 1

Simplify the left side.

Simplify $(x + 5) (x - 3)$ .

Expand $(x + 5) (x - 3)$ using the FOIL Method.

Apply the distributive property.

$x (x - 3) + 5 (x - 3) = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 3 + 5 (x - 3) = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 3 + 5 x + 5 \cdot - 3 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 3 + 5 x + 5 \cdot - 3 = 0$

Move $- 3$ to the left of $x$ .

$x^{2} - 3 \cdot x + 5 x + 5 \cdot - 3 = 0$

Multiply $5$ by $- 3$ .

$x^{2} - 3 x + 5 x - 15 = 0$

Add $- 3 x$ and $5 x$ .

$x^{2} + 2 x - 15 = 0$

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 $a = 1$ , $b = 2$ , and $c = - 15$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 15)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 15$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4$ by $- 15$ .

$x = \frac{- 2 \pm \sqrt{4 + 60}}{2 \cdot 1}$

Add $4$ and $60$ .

$x = \frac{- 2 \pm \sqrt{64}}{2 \cdot 1}$

Rewrite $64$ as $8^{2}$ .

$x = \frac{- 2 \pm \sqrt{8^{2}}}{2 \cdot 1}$

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

$x = \frac{- 2 \pm 8}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 8}{2}$

Simplify $\frac{- 2 \pm 8}{2}$ .

$x = - 1 \pm 4$

Step 5

The final answer is the combination of both solutions.

$x = 3, - 5$

Solve Using the Quadratic Formula (x+5)(x-3)=0

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