Solve 0=2n^2+3n-340

$$0 = 2 n ^ { 2 } + 3 n - 340$$

Solve for n

$$n = \frac{\sqrt{2729} - 3}{4} \approx 12.309957887$$

$$n=\frac{-\sqrt{2729}-3}{4}\approx -13.809957887$$

Steps Using the Quadratic Formula

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

$$2n^{2}+3n-340=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 3 for b, and -340 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-3±\sqrt{3^{2}-4\times 2\left(-340\right)}}{2\times 2}$$

Square 3.

$$n=\frac{-3±\sqrt{9-4\times 2\left(-340\right)}}{2\times 2}$$

Multiply -4 times 2.

$$n=\frac{-3±\sqrt{9-8\left(-340\right)}}{2\times 2}$$

Multiply -8 times -340.

$$n=\frac{-3±\sqrt{9+2720}}{2\times 2}$$

Add 9 to 2720.

$$n=\frac{-3±\sqrt{2729}}{2\times 2}$$

Multiply 2 times 2.

$$n=\frac{-3±\sqrt{2729}}{4}$$

Now solve the equation n=\frac{-3±\sqrt{2729}}{4} when ± is plus. Add -3 to \sqrt{2729}.

$$n=\frac{\sqrt{2729}-3}{4}$$

Now solve the equation n=\frac{-3±\sqrt{2729}}{4} when ± is minus. Subtract \sqrt{2729} from -3.

$$n=\frac{-\sqrt{2729}-3}{4}$$

The equation is now solved.

$$n=\frac{\sqrt{2729}-3}{4} n=\frac{-\sqrt{2729}-3}{4}$$

Steps for Completing the Square

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

$$2n^{2}+3n-340=0$$

Add 340 to both sides. Anything plus zero gives itself.

$$2n^{2}+3n=340$$

Divide both sides by 2.

$$\frac{2n^{2}+3n}{2}=\frac{340}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$n^{2}+\frac{3}{2}n=\frac{340}{2}$$

Divide 340 by 2.

$$n^{2}+\frac{3}{2}n=170$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{3}{2}n+\left(\frac{3}{4}\right)^{2}=170+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{3}{2}n+\frac{9}{16}=170+\frac{9}{16}$$

Add 170 to \frac{9}{16}.

$$n^{2}+\frac{3}{2}n+\frac{9}{16}=\frac{2729}{16}$$

Factor n^{2}+\frac{3}{2}n+\frac{9}{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}.

$$\left(n+\frac{3}{4}\right)^{2}=\frac{2729}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{3}{4}\right)^{2}}=\sqrt{\frac{2729}{16}}$$

Simplify.

$$n+\frac{3}{4}=\frac{\sqrt{2729}}{4} n+\frac{3}{4}=-\frac{\sqrt{2729}}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$n=\frac{\sqrt{2729}-3}{4} n=\frac{-\sqrt{2729}-3}{4}$$