Solve 5+x^2=1

$$5 + x ^ { 2 } = 1$$

Solve for x (complex solution)

$$x=-2i$$

$$x=2i$$

Steps by Finding Square Root

Subtract 5 from both sides.

$$x^{2}=1-5$$

Subtract 5 from 1 to get -4.

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

The equation is now solved.

$$x=2i x=-2i$$

Steps Using the Quadratic Formula

Subtract 1 from both sides.

$$5+x^{2}-1=0$$

Subtract 1 from 5 to get 4.

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

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

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

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}.

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

Square 0.

$$x=\frac{0±\sqrt{-4\times 4}}{2}$$

Multiply -4 times 4.

$$x=\frac{0±\sqrt{-16}}{2}$$

Take the square root of -16.

$$x=\frac{0±4i}{2}$$

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

$$x=2i$$

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

$$x=-2i$$

The equation is now solved.

$$x=2i x=-2i$$