Solve Using the Quadratic Formula 11x^2+33x=0

$11 x^{2} + 33 x = 0$

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 $a = 11$ , $b = 33$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{- 33 \pm \sqrt{33^{2} - 4 \cdot (11 \cdot 0)}}{2 \cdot 11}$

Step 3

Simplify.

Simplify the numerator.

Raise $33$ to the power of $2$ .

$x = \frac{- 33 \pm \sqrt{1089 - 4 \cdot 11 \cdot 0}}{2 \cdot 11}$

Multiply $- 4 \cdot 11 \cdot 0$ .

Multiply $- 4$ by $11$ .

$x = \frac{- 33 \pm \sqrt{1089 - 44 \cdot 0}}{2 \cdot 11}$

Multiply $- 44$ by $0$ .

$x = \frac{- 33 \pm \sqrt{1089 + 0}}{2 \cdot 11}$

Add $1089$ and $0$ .

$x = \frac{- 33 \pm \sqrt{1089}}{2 \cdot 11}$

Rewrite $1089$ as $33^{2}$ .

$x = \frac{- 33 \pm \sqrt{33^{2}}}{2 \cdot 11}$

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

$x = \frac{- 33 \pm 33}{2 \cdot 11}$

Multiply $2$ by $11$ .

$x = \frac{- 33 \pm 33}{22}$

Simplify $\frac{- 33 \pm 33}{22}$ .

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

Step 4

The final answer is the combination of both solutions.

$x = 0, - 3$