Solve n^2=9

$$n ^ { 2 } = 9$$

Solve for n

$$n=3$$

$$n=-3$$

Steps Using Difference of Squares

Subtract 9 from both sides.

$$n^{2}-9=0$$

Consider n^{2}-9. Rewrite n^{2}-9 as n^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(n-3\right)\left(n+3\right)=0$$

To find equation solutions, solve n-3=0 and n+3=0.

$$n=3 n=-3$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$n=3 n=-3$$

Steps Using the Quadratic Formula

Subtract 9 from both sides.

$$n^{2}-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{0±\sqrt{0^{2}-4\left(-9\right)}}{2}$$

Square 0.

$$n=\frac{0±\sqrt{-4\left(-9\right)}}{2}$$

Multiply -4 times -9.

$$n=\frac{0±\sqrt{36}}{2}$$

Take the square root of 36.

$$n=\frac{0±6}{2}$$

Now solve the equation n=\frac{0±6}{2} when ± is plus. Divide 6 by 2.

$$n=3$$

Now solve the equation n=\frac{0±6}{2} when ± is minus. Divide -6 by 2.

$$n=-3$$

The equation is now solved.

$$n=3 n=-3$$