Solve n^2+3=4

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

Solve for n

$$n=1$$

$$n=-1$$

Steps Using Difference of Squares

Subtract 4 from both sides.

$$n^{2}+3-4=0$$

Subtract 4 from 3 to get -1.

$$n^{2}-1=0$$

Consider n^{2}-1. Rewrite n^{2}-1 as n^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(n-1\right)\left(n+1\right)=0$$

To find equation solutions, solve n-1=0 and n+1=0.

$$n=1 n=-1$$

Steps by Finding Square Root

Subtract 3 from both sides.

$$n^{2}=4-3$$

Subtract 3 from 4 to get 1.

$$n^{2}=1$$

Take the square root of both sides of the equation.

$$n=1 n=-1$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

$$n^{2}+3-4=0$$

Subtract 4 from 3 to get -1.

$$n^{2}-1=0$$

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

$$n=\frac{0±\sqrt{0^{2}-4\left(-1\right)}}{2}$$

Square 0.

$$n=\frac{0±\sqrt{-4\left(-1\right)}}{2}$$

Multiply -4 times -1.

$$n=\frac{0±\sqrt{4}}{2}$$

Take the square root of 4.

$$n=\frac{0±2}{2}$$

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

$$n=1$$

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

$$n=-1$$

The equation is now solved.

$$n=1 n=-1$$