Solve n^2=4

$$n ^ { 2 } = 4$$

Solve for n

$$n=2$$

$$n=-2$$

Steps Using Difference of Squares

Subtract 4 from both sides.

$$n^{2}-4=0$$

Consider n^{2}-4. Rewrite n^{2}-4 as n^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(n-2\right)\left(n+2\right)=0$$

To find equation solutions, solve n-2=0 and n+2=0.

$$n=2 n=-2$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$n=2 n=-2$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

$$n^{2}-4=0$$

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

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

Square 0.

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

Multiply -4 times -4.

$$n=\frac{0±\sqrt{16}}{2}$$

Take the square root of 16.

$$n=\frac{0±4}{2}$$

Now solve the equation n=\frac{0±4}{2} when ± is plus. Divide 4 by 2.

$$n=2$$

Now solve the equation n=\frac{0±4}{2} when ± is minus. Divide -4 by 2.

$$n=-2$$

The equation is now solved.

$$n=2 n=-2$$