Solve 2+2a^2/4=a^2

$$2 + \frac { 2 a ^ { 2 } } { 4 } = a ^ { 2 }$$

Solve for a

$$a=2$$

$$a=-2$$

Steps by Finding Square Root

Multiply both sides of the equation by 4.

$$8+2a^{2}=4a^{2}$$

Subtract 4a^{2} from both sides.

$$8+2a^{2}-4a^{2}=0$$

Combine 2a^{2} and -4a^{2} to get -2a^{2}.

$$8-2a^{2}=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$-2a^{2}=-8$$

Divide both sides by -2.

$$a^{2}=\frac{-8}{-2}$$

Divide -8 by -2 to get 4.

$$a^{2}=4$$

Take the square root of both sides of the equation.

$$a=2 a=-2$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$8+2a^{2}=4a^{2}$$

Subtract 4a^{2} from both sides.

$$8+2a^{2}-4a^{2}=0$$

Combine 2a^{2} and -4a^{2} to get -2a^{2}.

$$8-2a^{2}=0$$

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.

$$-2a^{2}+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 0 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{0±\sqrt{0^{2}-4\left(-2\right)\times 8}}{2\left(-2\right)}$$

Square 0.

$$a=\frac{0±\sqrt{-4\left(-2\right)\times 8}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$a=\frac{0±\sqrt{8\times 8}}{2\left(-2\right)}$$

Multiply 8 times 8.

$$a=\frac{0±\sqrt{64}}{2\left(-2\right)}$$

Take the square root of 64.

$$a=\frac{0±8}{2\left(-2\right)}$$

Multiply 2 times -2.

$$a=\frac{0±8}{-4}$$

Now solve the equation a=\frac{0±8}{-4} when ± is plus. Divide 8 by -4.

$$a=-2$$

Now solve the equation a=\frac{0±8}{-4} when ± is minus. Divide -8 by -4.

$$a=2$$

The equation is now solved.

$$a=-2 a=2$$