Solve 2n^2+3n-2+3=9

$$2 n ^ { 2 } + 3 n - 2 + 3 = 9$$

Solve for n

$$n = \frac{\sqrt{73} - 3}{4} \approx 1.386000936$$

$$n=\frac{-\sqrt{73}-3}{4}\approx -2.886000936$$

Steps Using the Quadratic Formula

Add -2 and 3 to get 1.

$$2n^{2}+3n+1=9$$

Subtract 9 from both sides.

$$2n^{2}+3n+1-9=0$$

Subtract 9 from 1 to get -8.

$$2n^{2}+3n-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 3 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-3±\sqrt{3^{2}-4\times 2\left(-8\right)}}{2\times 2}$$

Square 3.

$$n=\frac{-3±\sqrt{9-4\times 2\left(-8\right)}}{2\times 2}$$

Multiply -4 times 2.

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

Multiply -8 times -8.

$$n=\frac{-3±\sqrt{9+64}}{2\times 2}$$

Add 9 to 64.

$$n=\frac{-3±\sqrt{73}}{2\times 2}$$

Multiply 2 times 2.

$$n=\frac{-3±\sqrt{73}}{4}$$

Now solve the equation n=\frac{-3±\sqrt{73}}{4} when ± is plus. Add -3 to \sqrt{73}.

$$n=\frac{\sqrt{73}-3}{4}$$

Now solve the equation n=\frac{-3±\sqrt{73}}{4} when ± is minus. Subtract \sqrt{73} from -3.

$$n=\frac{-\sqrt{73}-3}{4}$$

The equation is now solved.

$$n=\frac{\sqrt{73}-3}{4} n=\frac{-\sqrt{73}-3}{4}$$

Steps for Completing the Square

Add -2 and 3 to get 1.

$$2n^{2}+3n+1=9$$

Subtract 1 from both sides.

$$2n^{2}+3n=9-1$$

Subtract 1 from 9 to get 8.

$$2n^{2}+3n=8$$

Divide both sides by 2.

$$\frac{2n^{2}+3n}{2}=\frac{8}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$n^{2}+\frac{3}{2}n=\frac{8}{2}$$

Divide 8 by 2.

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

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.

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

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

$$n^{2}+\frac{3}{2}n+\frac{9}{16}=4+\frac{9}{16}$$

Add 4 to \frac{9}{16}.

$$n^{2}+\frac{3}{2}n+\frac{9}{16}=\frac{73}{16}$$

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}.

$$\left(n+\frac{3}{4}\right)^{2}=\frac{73}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{3}{4}\right)^{2}}=\sqrt{\frac{73}{16}}$$

Simplify.

$$n+\frac{3}{4}=\frac{\sqrt{73}}{4} n+\frac{3}{4}=-\frac{\sqrt{73}}{4}$$

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

$$n=\frac{\sqrt{73}-3}{4} n=\frac{-\sqrt{73}-3}{4}$$