Solve 20x^2=-25x

$$20 x ^ { 2 } = - 25 x$$

Solve for x

$$x = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

$$x=0$$

Steps Using Factoring

Add 25x to both sides.

$$20x^{2}+25x=0$$

Factor out x.

$$x\left(20x+25\right)=0$$

To find equation solutions, solve x=0 and 20x+25=0.

$$x=0 x=-\frac{5}{4}$$

Steps Using the Quadratic Formula

Add 25x to both sides.

$$20x^{2}+25x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, 25 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}}}{2\times 20}$$

Take the square root of 25^{2}.

$$x=\frac{-25±25}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-25±25}{40}$$

Now solve the equation x=\frac{-25±25}{40} when ± is plus. Add -25 to 25.

$$x=\frac{0}{40}$$

Divide 0 by 40.

$$x=0$$

Now solve the equation x=\frac{-25±25}{40} when ± is minus. Subtract 25 from -25.

$$x=-\frac{50}{40}$$

Reduce the fraction \frac{-50}{40} to lowest terms by extracting and canceling out 10.

$$x=-\frac{5}{4}$$

The equation is now solved.

$$x=0 x=-\frac{5}{4}$$

Steps for Completing the Square

Add 25x to both sides.

$$20x^{2}+25x=0$$

Divide both sides by 20.

$$\frac{20x^{2}+25x}{20}=\frac{0}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}+\frac{25}{20}x=\frac{0}{20}$$

Reduce the fraction \frac{25}{20} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{5}{4}x=\frac{0}{20}$$

Divide 0 by 20.

$$x^{2}+\frac{5}{4}x=0$$

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{25}{64}$$

Factor x^{2}+\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{8}\right)^{2}=\frac{25}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Simplify.

$$x+\frac{5}{8}=\frac{5}{8} x+\frac{5}{8}=-\frac{5}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$x=0 x=-\frac{5}{4}$$