Solve n^2-2n-8=(n

$$n ^ { 2 } - 2 n - 8 = ( n$$

Solve for n

$$n = \frac{\sqrt{41} + 3}{2} \approx 4.701562119$$

$$n=\frac{3-\sqrt{41}}{2}\approx -1.701562119$$

Steps Using the Quadratic Formula

Subtract n from both sides.

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

Combine -2n and -n to get -3n.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -3.

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

Multiply -4 times -8.

$$n=\frac{-\left(-3\right)±\sqrt{9+32}}{2}$$

Add 9 to 32.

$$n=\frac{-\left(-3\right)±\sqrt{41}}{2}$$

The opposite of -3 is 3.

$$n=\frac{3±\sqrt{41}}{2}$$

Now solve the equation n=\frac{3±\sqrt{41}}{2} when ± is plus. Add 3 to \sqrt{41}.

$$n=\frac{\sqrt{41}+3}{2}$$

Now solve the equation n=\frac{3±\sqrt{41}}{2} when ± is minus. Subtract \sqrt{41} from 3.

$$n=\frac{3-\sqrt{41}}{2}$$

The equation is now solved.

$$n=\frac{\sqrt{41}+3}{2} n=\frac{3-\sqrt{41}}{2}$$

Steps for Completing the Square

Subtract n from both sides.

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

Combine -2n and -n to get -3n.

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

Add 8 to both sides. Anything plus zero gives itself.

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

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

$$n^{2}-3n+\left(-\frac{3}{2}\right)^{2}=8+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-3n+\frac{9}{4}=8+\frac{9}{4}$$

Add 8 to \frac{9}{4}.

$$n^{2}-3n+\frac{9}{4}=\frac{41}{4}$$

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

$$\left(n-\frac{3}{2}\right)^{2}=\frac{41}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{3}{2}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Simplify.

$$n-\frac{3}{2}=\frac{\sqrt{41}}{2} n-\frac{3}{2}=-\frac{\sqrt{41}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$n=\frac{\sqrt{41}+3}{2} n=\frac{3-\sqrt{41}}{2}$$