Solve 3n^2-13=3n

$$3 n ^ { 2 } - 13 = 3 n$$

Solve for n

$$n=\frac{\sqrt{165}}{6}+\frac{1}{2}\approx 2.640872096$$

$$n=-\frac{\sqrt{165}}{6}+\frac{1}{2}\approx -1.640872096$$

Steps Using the Quadratic Formula

Subtract 3n from both sides.

$$3n^{2}-13-3n=0$$

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.

$$3n^{2}-3n-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -3 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 3\left(-13\right)}}{2\times 3}$$

Square -3.

$$n=\frac{-\left(-3\right)±\sqrt{9-4\times 3\left(-13\right)}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{-\left(-3\right)±\sqrt{9-12\left(-13\right)}}{2\times 3}$$

Multiply -12 times -13.

$$n=\frac{-\left(-3\right)±\sqrt{9+156}}{2\times 3}$$

Add 9 to 156.

$$n=\frac{-\left(-3\right)±\sqrt{165}}{2\times 3}$$

The opposite of -3 is 3.

$$n=\frac{3±\sqrt{165}}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{3±\sqrt{165}}{6}$$

Now solve the equation n=\frac{3±\sqrt{165}}{6} when ± is plus. Add 3 to \sqrt{165}.

$$n=\frac{\sqrt{165}+3}{6}$$

Divide 3+\sqrt{165} by 6.

$$n=\frac{\sqrt{165}}{6}+\frac{1}{2}$$

Now solve the equation n=\frac{3±\sqrt{165}}{6} when ± is minus. Subtract \sqrt{165} from 3.

$$n=\frac{3-\sqrt{165}}{6}$$

Divide 3-\sqrt{165} by 6.

$$n=-\frac{\sqrt{165}}{6}+\frac{1}{2}$$

The equation is now solved.

$$n=\frac{\sqrt{165}}{6}+\frac{1}{2} n=-\frac{\sqrt{165}}{6}+\frac{1}{2}$$

Steps for Completing the Square

Subtract 3n from both sides.

$$3n^{2}-13-3n=0$$

Add 13 to both sides. Anything plus zero gives itself.

$$3n^{2}-3n=13$$

Divide both sides by 3.

$$\frac{3n^{2}-3n}{3}=\frac{13}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$n^{2}+\left(-\frac{3}{3}\right)n=\frac{13}{3}$$

Divide -3 by 3.

$$n^{2}-n=\frac{13}{3}$$

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.

$$n^{2}-n+\left(-\frac{1}{2}\right)^{2}=\frac{13}{3}+\left(-\frac{1}{2}\right)^{2}$$

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

$$n^{2}-n+\frac{1}{4}=\frac{13}{3}+\frac{1}{4}$$

Add \frac{13}{3} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-n+\frac{1}{4}=\frac{55}{12}$$

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}.

$$\left(n-\frac{1}{2}\right)^{2}=\frac{55}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{1}{2}\right)^{2}}=\sqrt{\frac{55}{12}}$$

Simplify.

$$n-\frac{1}{2}=\frac{\sqrt{165}}{6} n-\frac{1}{2}=-\frac{\sqrt{165}}{6}$$

Add \frac{1}{2} to both sides of the equation.

$$n=\frac{\sqrt{165}}{6}+\frac{1}{2} n=-\frac{\sqrt{165}}{6}+\frac{1}{2}$$