Solve 4-x^2/2=x^2/2

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

Solve for x

$$x=2$$

$$x=-2$$

Steps by Finding Square Root

Multiply both sides of the equation by 2.

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

Subtract x^{2} from both sides.

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

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

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

Combine -x^{2} and -x^{2} to get -2x^{2}.

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

Divide both sides by -2.

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

Divide -8 by -2 to get 4.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

Subtract x^{2} from both sides.

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

Combine -x^{2} and -x^{2} to get -2x^{2}.

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

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.

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

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}.

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

Square 0.

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

Multiply -4 times -2.

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

Multiply 8 times 8.

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

Take the square root of 64.

$$x=\frac{0±8}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{0±8}{-4}$$

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

$$x=-2$$

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

$$x=2$$

The equation is now solved.

$$x=-2 x=2$$