Solve Using the Quadratic Formula x^2-5x-36=0

$x^{2} - 5 x - 36 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 1$ , $b = - 5$ , and $c = - 36$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (1 \cdot - 36)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 36}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 36$ .

Multiply $- 4$ by $1$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot - 36}}{2 \cdot 1}$

Multiply $- 4$ by $- 36$ .

$x = \frac{5 \pm \sqrt{25 + 144}}{2 \cdot 1}$

Add $25$ and $144$ .

$x = \frac{5 \pm \sqrt{169}}{2 \cdot 1}$

Rewrite $169$ as $13^{2}$ .

$x = \frac{5 \pm \sqrt{13^{2}}}{2 \cdot 1}$

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

$x = \frac{5 \pm 13}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{5 \pm 13}{2}$

Step 4

The final answer is the combination of both solutions.

$x = 9, - 4$