Solve Using the Quadratic Formula (3x)^2-12(3x)+27=0

${(3 x)}^{2} - 12 (3 x) + 27 = 0$

Step 1

Simplify the left side.

Simplify each term.

Apply the product rule to $3 x$ .

$3^{2} x^{2} - 12 (3 x) + 27 = 0$

Raise $3$ to the power of $2$ .

$9 x^{2} - 12 (3 x) + 27 = 0$

Multiply $3$ by $- 12$ .

$9 x^{2} - 36 x + 27 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 9$ , $b = - 36$ , and $c = 27$ into the quadratic formula and solve for $x$ .

$\frac{36 \pm \sqrt{{(- 36)}^{2} - 4 \cdot (9 \cdot 27)}}{2 \cdot 9}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 36$ to the power of $2$ .

$x = \frac{36 \pm \sqrt{1296 - 4 \cdot 9 \cdot 27}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot 27$ .

Multiply $- 4$ by $9$ .

$x = \frac{36 \pm \sqrt{1296 - 36 \cdot 27}}{2 \cdot 9}$

Multiply $- 36$ by $27$ .

$x = \frac{36 \pm \sqrt{1296 - 972}}{2 \cdot 9}$

Subtract $972$ from $1296$ .

$x = \frac{36 \pm \sqrt{324}}{2 \cdot 9}$

Rewrite $324$ as $18^{2}$ .

$x = \frac{36 \pm \sqrt{18^{2}}}{2 \cdot 9}$

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

$x = \frac{36 \pm 18}{2 \cdot 9}$

Multiply $2$ by $9$ .

$x = \frac{36 \pm 18}{18}$

Simplify $\frac{36 \pm 18}{18}$ .

$x = 2 \pm 1$

Step 5

The final answer is the combination of both solutions.

$x = 3, 1$