Solve Using the Quadratic Formula (x-4)(x+5)=10

$(x - 4) (x + 5) = 10$

Step 1

Move all terms to the left side of the equation and simplify.

Simplify the left side.

Simplify $(x - 4) (x + 5)$ .

Expand $(x - 4) (x + 5)$ using the FOIL Method.

Apply the distributive property.

$x (x + 5) - 4 (x + 5) = 10$

Apply the distributive property.

$x \cdot x + x \cdot 5 - 4 (x + 5) = 10$

Apply the distributive property.

$x \cdot x + x \cdot 5 - 4 x - 4 \cdot 5 = 10$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 5 - 4 x - 4 \cdot 5 = 10$

Move $5$ to the left of $x$ .

$x^{2} + 5 \cdot x - 4 x - 4 \cdot 5 = 10$

Multiply $- 4$ by $5$ .

$x^{2} + 5 x - 4 x - 20 = 10$

Subtract $4 x$ from $5 x$ .

$x^{2} + x - 20 = 10$

Subtract $10$ from both sides of the equation.

$x^{2} + x - 20 - 10 = 0$

Subtract $10$ from $- 20$ .

$x^{2} + x - 30 = 0$

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 $c = - 30$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot - 30)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 30}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 30$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 30}}{2 \cdot 1}$

Multiply $- 4$ by $- 30$ .

$x = \frac{- 1 \pm \sqrt{1 + 120}}{2 \cdot 1}$

Add $1$ and $120$ .

$x = \frac{- 1 \pm \sqrt{121}}{2 \cdot 1}$

Rewrite $121$ as $11^{2}$ .

$x = \frac{- 1 \pm \sqrt{11^{2}}}{2 \cdot 1}$

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

$x = \frac{- 1 \pm 11}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm 11}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 5, - 6$

Solve Using the Quadratic Formula (x-4)(x+5)=10

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