Solve x^2+16=4

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

Solve for x (complex solution)

$$x=-2\sqrt{3}i\approx -0-3.464101615i$$

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

Steps by Finding Square Root

Subtract 16 from both sides.

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

Subtract 16 from 4 to get -12.

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

The equation is now solved.

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

Steps Using the Quadratic Formula

Subtract 4 from both sides.

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

Subtract 4 from 16 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}.

$$x=\frac{0±\sqrt{0^{2}-4\times 12}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 12}}{2}$$

Multiply -4 times 12.

$$x=\frac{0±\sqrt{-48}}{2}$$

Take the square root of -48.

$$x=\frac{0±4\sqrt{3}i}{2}$$

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

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

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

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

The equation is now solved.

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