Solve 30=2n^2-n

$$30 = 2 n ^ { 2 } - n$$

Solve for n

$$n = \frac{\sqrt{241} + 1}{4} \approx 4.131043674$$

$$n=\frac{1-\sqrt{241}}{4}\approx -3.631043674$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$2n^{2}-n=30$$

Subtract 30 from both sides.

$$2n^{2}-n-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-30\right)}}{2\times 2}$$

Multiply -4 times 2.

$$n=\frac{-\left(-1\right)±\sqrt{1-8\left(-30\right)}}{2\times 2}$$

Multiply -8 times -30.

$$n=\frac{-\left(-1\right)±\sqrt{1+240}}{2\times 2}$$

Add 1 to 240.

$$n=\frac{-\left(-1\right)±\sqrt{241}}{2\times 2}$$

The opposite of -1 is 1.

$$n=\frac{1±\sqrt{241}}{2\times 2}$$

Multiply 2 times 2.

$$n=\frac{1±\sqrt{241}}{4}$$

Now solve the equation n=\frac{1±\sqrt{241}}{4} when ± is plus. Add 1 to \sqrt{241}.

$$n=\frac{\sqrt{241}+1}{4}$$

Now solve the equation n=\frac{1±\sqrt{241}}{4} when ± is minus. Subtract \sqrt{241} from 1.

$$n=\frac{1-\sqrt{241}}{4}$$

The equation is now solved.

$$n=\frac{\sqrt{241}+1}{4} n=\frac{1-\sqrt{241}}{4}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$2n^{2}-n=30$$

Divide both sides by 2.

$$\frac{2n^{2}-n}{2}=\frac{30}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$n^{2}-\frac{1}{2}n=\frac{30}{2}$$

Divide 30 by 2.

$$n^{2}-\frac{1}{2}n=15$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{1}{2}n+\left(-\frac{1}{4}\right)^{2}=15+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{1}{2}n+\frac{1}{16}=15+\frac{1}{16}$$

Add 15 to \frac{1}{16}.

$$n^{2}-\frac{1}{2}n+\frac{1}{16}=\frac{241}{16}$$

Factor n^{2}-\frac{1}{2}n+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{1}{4}\right)^{2}=\frac{241}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{1}{4}\right)^{2}}=\sqrt{\frac{241}{16}}$$

Simplify.

$$n-\frac{1}{4}=\frac{\sqrt{241}}{4} n-\frac{1}{4}=-\frac{\sqrt{241}}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$n=\frac{\sqrt{241}+1}{4} n=\frac{1-\sqrt{241}}{4}$$