Solve Using the Quadratic Formula b^2-12b-64=0

$b^{2} - 12 b - 64 = 0$

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 $a = 1$ , $b = - 12$ , and $c = - 64$ into the quadratic formula and solve for $b$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (1 \cdot - 64)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 12$ to the power of $2$ .

$b = \frac{12 \pm \sqrt{144 - 4 \cdot 1 \cdot - 64}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 64$ .

Multiply $- 4$ by $1$ .

$b = \frac{12 \pm \sqrt{144 - 4 \cdot - 64}}{2 \cdot 1}$

Multiply $- 4$ by $- 64$ .

$b = \frac{12 \pm \sqrt{144 + 256}}{2 \cdot 1}$

Add $144$ and $256$ .

$b = \frac{12 \pm \sqrt{400}}{2 \cdot 1}$

Rewrite $400$ as $20^{2}$ .

$b = \frac{12 \pm \sqrt{20^{2}}}{2 \cdot 1}$

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

$b = \frac{12 \pm 20}{2 \cdot 1}$

Multiply $2$ by $1$ .

$b = \frac{12 \pm 20}{2}$

Simplify $\frac{12 \pm 20}{2}$ .

$b = 6 \pm 10$

Step 4

The final answer is the combination of both solutions.

$b = 16, - 4$