Solve 2u^2+3=1

$$2 u ^ { 2 } + 3 = 1$$

Solve for u

$$u=-i$$

$$u=i$$

Steps by Finding Square Root

Subtract 3 from both sides.

$$2u^{2}=1-3$$

Subtract 3 from 1 to get -2.

$$2u^{2}=-2$$

Divide both sides by 2.

$$u^{2}=\frac{-2}{2}$$

Divide -2 by 2 to get -1.

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

The equation is now solved.

$$u=i u=-i$$

Steps Using the Quadratic Formula

Subtract 1 from both sides.

$$2u^{2}+3-1=0$$

Subtract 1 from 3 to get 2.

$$2u^{2}+2=0$$

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

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

Square 0.

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

Multiply -4 times 2.

$$u=\frac{0±\sqrt{-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$u=\frac{0±\sqrt{-16}}{2\times 2}$$

Take the square root of -16.

$$u=\frac{0±4i}{2\times 2}$$

Multiply 2 times 2.

$$u=\frac{0±4i}{4}$$

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

$$u=i$$

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

$$u=-i$$

The equation is now solved.

$$u=i u=-i$$