Solve 3n^2=-3n

$$3 n ^ { 2 } = - 3 n$$

Solve for n

$$n=-1$$

$$n=0$$

Steps Using the Quadratic Formula

Add 3n to both sides.

$$3n^{2}+3n=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-3±\sqrt{3^{2}}}{2\times 3}$$

Take the square root of 3^{2}.

$$n=\frac{-3±3}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{-3±3}{6}$$

Now solve the equation n=\frac{-3±3}{6} when ± is plus. Add -3 to 3.

$$n=\frac{0}{6}$$

Divide 0 by 6.

$$n=0$$

Now solve the equation n=\frac{-3±3}{6} when ± is minus. Subtract 3 from -3.

$$n=-\frac{6}{6}$$

Divide -6 by 6.

$$n=-1$$

The equation is now solved.

$$n=0 n=-1$$

Steps for Completing the Square

Add 3n to both sides.

$$3n^{2}+3n=0$$

Divide both sides by 3.

$$\frac{3n^{2}+3n}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$n^{2}+\frac{3}{3}n=\frac{0}{3}$$

Divide 3 by 3.

$$n^{2}+n=\frac{0}{3}$$

Divide 0 by 3.

$$n^{2}+n=0$$

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+n+\frac{1}{4}=\frac{1}{4}$$

Factor n^{2}+n+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{1}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$n+\frac{1}{2}=\frac{1}{2} n+\frac{1}{2}=-\frac{1}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$n=0 n=-1$$