Solve Using the Quadratic Formula 9x(x-4)=0

$9 x (x - 4) = 0$

Step 1

Simplify the left side.

Simplify $9 x (x - 4)$ .

Apply the distributive property.

$9 x \cdot x + 9 x \cdot - 4 = 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$9 (x \cdot x) + 9 x \cdot - 4 = 0$

Multiply $x$ by $x$ .

$9 x^{2} + 9 x \cdot - 4 = 0$

Multiply $- 4$ by $9$ .

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

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 $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{36 \pm \sqrt{{(- 36)}^{2} - 4 \cdot (9 \cdot 0)}}{2 \cdot 9}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{36 \pm \sqrt{1296 - 4 \cdot 9 \cdot 0}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot 0$ .

Multiply $- 4$ by $9$ .

$x = \frac{36 \pm \sqrt{1296 - 36 \cdot 0}}{2 \cdot 9}$

Multiply $- 36$ by $0$ .

$x = \frac{36 \pm \sqrt{1296 + 0}}{2 \cdot 9}$

Add $1296$ and $0$ .

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

Rewrite $1296$ as $36^{2}$ .

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

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

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

Multiply $2$ by $9$ .

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

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

$x = 2 \pm 2$

Step 5

The final answer is the combination of both solutions.

$x = 4, 0$