Solve 30r^2=-25r

$$30 r ^ { 2 } = - 25 r$$

Solve for r

$$r=-\frac{5}{6}\approx -0.833333333$$

$$r=0$$

Steps Using Factoring

Add 25r to both sides.

$$30r^{2}+25r=0$$

Factor out r.

$$r\left(30r+25\right)=0$$

To find equation solutions, solve r=0 and 30r+25=0.

$$r=0 r=-\frac{5}{6}$$

Steps Using the Quadratic Formula

Add 25r to both sides.

$$30r^{2}+25r=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 30 for a, 25 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-25±\sqrt{25^{2}}}{2\times 30}$$

Take the square root of 25^{2}.

$$r=\frac{-25±25}{2\times 30}$$

Multiply 2 times 30.

$$r=\frac{-25±25}{60}$$

Now solve the equation r=\frac{-25±25}{60} when ± is plus. Add -25 to 25.

$$r=\frac{0}{60}$$

Divide 0 by 60.

$$r=0$$

Now solve the equation r=\frac{-25±25}{60} when ± is minus. Subtract 25 from -25.

$$r=-\frac{50}{60}$$

Reduce the fraction \frac{-50}{60} to lowest terms by extracting and canceling out 10.

$$r=-\frac{5}{6}$$

The equation is now solved.

$$r=0 r=-\frac{5}{6}$$

Steps for Completing the Square

Add 25r to both sides.

$$30r^{2}+25r=0$$

Divide both sides by 30.

$$\frac{30r^{2}+25r}{30}=\frac{0}{30}$$

Dividing by 30 undoes the multiplication by 30.

$$r^{2}+\frac{25}{30}r=\frac{0}{30}$$

Reduce the fraction \frac{25}{30} to lowest terms by extracting and canceling out 5.

$$r^{2}+\frac{5}{6}r=\frac{0}{30}$$

Divide 0 by 30.

$$r^{2}+\frac{5}{6}r=0$$

Divide \frac{5}{6}, the coefficient of the x term, by 2 to get \frac{5}{12}. Then add the square of \frac{5}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}+\frac{5}{6}r+\left(\frac{5}{12}\right)^{2}=\left(\frac{5}{12}\right)^{2}$$

Square \frac{5}{12} by squaring both the numerator and the denominator of the fraction.

$$r^{2}+\frac{5}{6}r+\frac{25}{144}=\frac{25}{144}$$

Factor r^{2}+\frac{5}{6}r+\frac{25}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(r+\frac{5}{12}\right)^{2}=\frac{25}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r+\frac{5}{12}\right)^{2}}=\sqrt{\frac{25}{144}}$$

Simplify.

$$r+\frac{5}{12}=\frac{5}{12} r+\frac{5}{12}=-\frac{5}{12}$$

Subtract \frac{5}{12} from both sides of the equation.

$$r=0 r=-\frac{5}{6}$$