Solve 4x^2+5=1

$$4 x ^ { 2 } + 5 = 1$$

Solve for x (complex solution)

$$x=-i$$

$$x=i$$

Steps by Finding Square Root

Subtract 5 from both sides.

$$4x^{2}=1-5$$

Subtract 5 from 1 to get -4.

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

Divide both sides by 4.

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

Divide -4 by 4 to get -1.

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

The equation is now solved.

$$x=i x=-i$$

Steps Using the Quadratic Formula

Subtract 1 from both sides.

$$4x^{2}+5-1=0$$

Subtract 1 from 5 to get 4.

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

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

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

Square 0.

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

Multiply -4 times 4.

$$x=\frac{0±\sqrt{-16\times 4}}{2\times 4}$$

Multiply -16 times 4.

$$x=\frac{0±\sqrt{-64}}{2\times 4}$$

Take the square root of -64.

$$x=\frac{0±8i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{0±8i}{8}$$

Now solve the equation x=\frac{0±8i}{8} when ± is plus.

$$x=i$$

Now solve the equation x=\frac{0±8i}{8} when ± is minus.

$$x=-i$$

The equation is now solved.

$$x=i x=-i$$