Solve 10(2x^2+2)=40

$$10 ( 2 x ^ { 2 } + 2 ) = 40$$

Solve for x

$$x=1$$

$$x=-1$$

Steps Using Difference of Squares

Divide both sides by 10.

$$2x^{2}+2=\frac{40}{10}$$

Divide 40 by 10 to get 4.

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

Subtract 4 from both sides.

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

Subtract 4 from 2 to get -2.

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

Divide both sides by 2.

$$x^{2}-1=0$$

Consider x^{2}-1. Rewrite x^{2}-1 as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-1\right)\left(x+1\right)=0$$

To find equation solutions, solve x-1=0 and x+1=0.

$$x=1 x=-1$$

Steps by Finding Square Root

Divide both sides by 10.

$$2x^{2}+2=\frac{40}{10}$$

Divide 40 by 10 to get 4.

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

Subtract 2 from both sides.

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

Subtract 2 from 4 to get 2.

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

Divide both sides by 2.

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

Divide 2 by 2 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Steps Using the Quadratic Formula

Divide both sides by 10.

$$2x^{2}+2=\frac{40}{10}$$

Divide 40 by 10 to get 4.

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

Subtract 4 from both sides.

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

Subtract 4 from 2 to get -2.

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

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}.

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

Square 0.

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

Multiply -4 times 2.

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

Multiply -8 times -2.

$$x=\frac{0±\sqrt{16}}{2\times 2}$$

Take the square root of 16.

$$x=\frac{0±4}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±4}{4}$$

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

$$x=1$$

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

$$x=-1$$

The equation is now solved.

$$x=1 x=-1$$