Solve Using the Quadratic Formula 4x^2-32x+64=0

$4 x^{2} - 32 x + 64 = 0$

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 $a = 4$ , $b = - 32$ , and $c = 64$ into the quadratic formula and solve for $x$ .

$\frac{32 \pm \sqrt{{(- 32)}^{2} - 4 \cdot (4 \cdot 64)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 32$ to the power of $2$ .

$x = \frac{32 \pm \sqrt{1024 - 4 \cdot 4 \cdot 64}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$x = \frac{32 \pm \sqrt{1024 - 16 \cdot 64}}{2 \cdot 4}$

Multiply $- 16$ by $64$ .

$x = \frac{32 \pm \sqrt{1024 - 1024}}{2 \cdot 4}$

Subtract $1024$ from $1024$ .

$x = \frac{32 \pm \sqrt{0}}{2 \cdot 4}$

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

$x = \frac{32 \pm \sqrt{0^{2}}}{2 \cdot 4}$

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

$x = \frac{32 \pm 0}{2 \cdot 4}$

$32$ plus or minus $0$ is $32$ .

$x = \frac{32}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{32}{8}$

Divide $32$ by $8$ .

$x = 4$

Step 4

The final answer is the combination of both solutions.

$x = 4$ Double roots