Solve 16=x^2+4

$$16 = x ^ { 2 } + 4$$

Solve for x

$$x=2\sqrt{3}\approx 3.464101615$$

$$x=-2\sqrt{3}\approx -3.464101615$$

Steps by Finding Square Root

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+4=16$$

Subtract 4 from both sides.

$$x^{2}=16-4$$

Subtract 4 from 16 to get 12.

$$x^{2}=12$$

Take the square root of both sides of the equation.

$$x=2\sqrt{3} x=-2\sqrt{3}$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+4=16$$

Subtract 16 from both sides.

$$x^{2}+4-16=0$$

Subtract 16 from 4 to get -12.

$$x^{2}-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-12\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$x=\frac{0±\sqrt{48}}{2}$$

Take the square root of 48.

$$x=\frac{0±4\sqrt{3}}{2}$$

Now solve the equation x=\frac{0±4\sqrt{3}}{2} when ± is plus.

$$x=2\sqrt{3}$$

Now solve the equation x=\frac{0±4\sqrt{3}}{2} when ± is minus.

$$x=-2\sqrt{3}$$

The equation is now solved.

$$x=2\sqrt{3} x=-2\sqrt{3}$$