Solve Using the Quadratic Formula a^2-12a+32=0

$a^{2} - 12 a + 32 = 0$

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 $c = 32$ into the quadratic formula and solve for $a$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (1 \cdot 32)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$a = \frac{12 \pm \sqrt{144 - 4 \cdot 1 \cdot 32}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 32$ .

Multiply $- 4$ by $1$ .

$a = \frac{12 \pm \sqrt{144 - 4 \cdot 32}}{2 \cdot 1}$

Multiply $- 4$ by $32$ .

$a = \frac{12 \pm \sqrt{144 - 128}}{2 \cdot 1}$

Subtract $128$ from $144$ .

$a = \frac{12 \pm \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$a = \frac{12 \pm \sqrt{4^{2}}}{2 \cdot 1}$

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

$a = \frac{12 \pm 4}{2 \cdot 1}$

Multiply $2$ by $1$ .

$a = \frac{12 \pm 4}{2}$

Simplify $\frac{12 \pm 4}{2}$ .

$a = 6 \pm 2$

Step 4

The final answer is the combination of both solutions.

$a = 8, 4$