Solve 2x^2+4=12

$$2 { x }^{ 2 } +4=12$$

Solve for x

$$x=2$$

$$x=-2$$

Steps Using Difference of Squares

Subtract 12 from both sides.

$$2x^{2}+4-12=0$$

Subtract 12 from 4 to get -8.

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

Divide both sides by 2.

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

Consider x^{2}-4. Rewrite x^{2}-4 as x^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-2\right)\left(x+2\right)=0$$

To find equation solutions, solve x-2=0 and x+2=0.

$$x=2 x=-2$$

Steps by Finding Square Root

Subtract 4 from both sides.

$$2x^{2}=12-4$$

Subtract 4 from 12 to get 8.

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

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

Subtract 12 from both sides.

$$2x^{2}+4-12=0$$

Subtract 12 from 4 to get -8.

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

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}.

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

Square 0.

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

Multiply -4 times 2.

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

Multiply -8 times -8.

$$x=\frac{0±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$x=\frac{0±8}{2\times 2}$$

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.

$$x=2 x=-2$$