Solve 18x^2-24x=0

$$18 x ^ { 2 } - 24 x = 0$$

Solve for x

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$x=0$$

Steps Using Factoring

Factor out x.

$$x\left(18x-24\right)=0$$

To find equation solutions, solve x=0 and 18x-24=0.

$$x=0 x=\frac{4}{3}$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$18x^{2}-24x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, -24 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}}}{2\times 18}$$

Take the square root of \left(-24\right)^{2}.

$$x=\frac{-\left(-24\right)±24}{2\times 18}$$

The opposite of -24 is 24.

$$x=\frac{24±24}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{24±24}{36}$$

Now solve the equation x=\frac{24±24}{36} when ± is plus. Add 24 to 24.

$$x=\frac{48}{36}$$

Reduce the fraction \frac{48}{36} to lowest terms by extracting and canceling out 12.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{24±24}{36} when ± is minus. Subtract 24 from 24.

$$x=\frac{0}{36}$$

Divide 0 by 36.

$$x=0$$

The equation is now solved.

$$x=\frac{4}{3} x=0$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$18x^{2}-24x=0$$

Divide both sides by 18.

$$\frac{18x^{2}-24x}{18}=\frac{0}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x^{2}+\left(-\frac{24}{18}\right)x=\frac{0}{18}$$

Reduce the fraction \frac{-24}{18} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{4}{3}x=\frac{0}{18}$$

Divide 0 by 18.

$$x^{2}-\frac{4}{3}x=0$$

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=\left(-\frac{2}{3}\right)^{2}$$

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{4}{9}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

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

Add \frac{2}{3} to both sides of the equation.

$$x=\frac{4}{3} x=0$$