Solve Using the Quadratic Formula x^2-16x+64=0

$x^{2} - 16 x + 64 = 0$

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 $a = 1$ , $b = - 16$ , and $c = 64$ into the quadratic formula and solve for $x$ .

$\frac{16 \pm \sqrt{{(- 16)}^{2} - 4 \cdot (1 \cdot 64)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 16$ to the power of $2$ .

$x = \frac{16 \pm \sqrt{256 - 4 \cdot 1 \cdot 64}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{16 \pm \sqrt{256 - 4 \cdot 64}}{2 \cdot 1}$

Multiply $- 4$ by $64$ .

$x = \frac{16 \pm \sqrt{256 - 256}}{2 \cdot 1}$

Subtract $256$ from $256$ .

$x = \frac{16 \pm \sqrt{0}}{2 \cdot 1}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{16 \pm \sqrt{0^{2}}}{2 \cdot 1}$

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

$x = \frac{16 \pm 0}{2 \cdot 1}$

$16$ plus or minus $0$ is $16$ .

$x = \frac{16}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{16}{2}$

Divide $16$ by $2$ .

$x = 8$

Step 4

The final answer is the combination of both solutions.

$x = 8$ Double roots