Solve 60x^2+24x=0

$$60 { x }^{ 2 } +24x=0$$

Solve for x

$$x=-\frac{2}{5}=-0.4$$

$$x=0$$

Steps Using Factoring

Factor out x.

$$x\left(60x+24\right)=0$$

To find equation solutions, solve x=0 and 60x+24=0.

$$x=0 x=-\frac{2}{5}$$

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.

$$60x^{2}+24x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 60 for a, 24 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-24±\sqrt{24^{2}}}{2\times 60}$$

Take the square root of 24^{2}.

$$x=\frac{-24±24}{2\times 60}$$

Multiply 2 times 60.

$$x=\frac{-24±24}{120}$$

Now solve the equation x=\frac{-24±24}{120} when ± is plus. Add -24 to 24.

$$x=\frac{0}{120}$$

Divide 0 by 120.

$$x=0$$

Now solve the equation x=\frac{-24±24}{120} when ± is minus. Subtract 24 from -24.

$$x=-\frac{48}{120}$$

Reduce the fraction \frac{-48}{120} to lowest terms by extracting and canceling out 24.

$$x=-\frac{2}{5}$$

The equation is now solved.

$$x=0 x=-\frac{2}{5}$$

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.

$$60x^{2}+24x=0$$

Divide both sides by 60.

$$\frac{60x^{2}+24x}{60}=\frac{0}{60}$$

Dividing by 60 undoes the multiplication by 60.

$$x^{2}+\frac{24}{60}x=\frac{0}{60}$$

Reduce the fraction \frac{24}{60} to lowest terms by extracting and canceling out 12.

$$x^{2}+\frac{2}{5}x=\frac{0}{60}$$

Divide 0 by 60.

$$x^{2}+\frac{2}{5}x=0$$

Divide \frac{2}{5}, the coefficient of the x term, by 2 to get \frac{1}{5}. Then add the square of \frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{5}x+\left(\frac{1}{5}\right)^{2}=\left(\frac{1}{5}\right)^{2}$$

Square \frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{1}{25}$$

Factor x^{2}+\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{5}\right)^{2}=\frac{1}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{5}\right)^{2}}=\sqrt{\frac{1}{25}}$$

Simplify.

$$x+\frac{1}{5}=\frac{1}{5} x+\frac{1}{5}=-\frac{1}{5}$$

Subtract \frac{1}{5} from both sides of the equation.

$$x=0 x=-\frac{2}{5}$$