Solve Using the Quadratic Formula x^2-15x+54=0

$x^{2} - 15 x + 54 = 0$

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 $a = 1$ , $b = - 15$ , and $c = 54$ into the quadratic formula and solve for $x$ .

$\frac{15 \pm \sqrt{{(- 15)}^{2} - 4 \cdot (1 \cdot 54)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 15$ to the power of $2$ .

$x = \frac{15 \pm \sqrt{225 - 4 \cdot 1 \cdot 54}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 54$ .

Multiply $- 4$ by $1$ .

$x = \frac{15 \pm \sqrt{225 - 4 \cdot 54}}{2 \cdot 1}$

Multiply $- 4$ by $54$ .

$x = \frac{15 \pm \sqrt{225 - 216}}{2 \cdot 1}$

Subtract $216$ from $225$ .

$x = \frac{15 \pm \sqrt{9}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{15 \pm \sqrt{3^{2}}}{2 \cdot 1}$

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

$x = \frac{15 \pm 3}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{15 \pm 3}{2}$

Step 4

The final answer is the combination of both solutions.

$x = 9, 6$