Solve 2(2x^2-32)=0

$$2 ( 2 x ^ { 2 } - 32 ) = 0$$

Solve for x

$$x=4$$

$$x=-4$$

Steps Using Difference of Squares

Divide both sides by 2. Zero divided by any non-zero number gives zero.

$$2x^{2}-32=0$$

Divide both sides by 2.

$$x^{2}-16=0$$

Consider x^{2}-16. Rewrite x^{2}-16 as x^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-4\right)\left(x+4\right)=0$$

To find equation solutions, solve x-4=0 and x+4=0.

$$x=4 x=-4$$

Steps by Finding Square Root

Divide both sides by 2. Zero divided by any non-zero number gives zero.

$$2x^{2}-32=0$$

Add 32 to both sides. Anything plus zero gives itself.

$$2x^{2}=32$$

Divide both sides by 2.

$$x^{2}=\frac{32}{2}$$

Divide 32 by 2 to get 16.

$$x^{2}=16$$

Take the square root of both sides of the equation.

$$x=4 x=-4$$

Steps Using the Quadratic Formula

Divide both sides by 2. Zero divided by any non-zero number gives zero.

$$2x^{2}-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 0 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\left(-32\right)}}{2\times 2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 2\left(-32\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{0±\sqrt{-8\left(-32\right)}}{2\times 2}$$

Multiply -8 times -32.

$$x=\frac{0±\sqrt{256}}{2\times 2}$$

Take the square root of 256.

$$x=\frac{0±16}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±16}{4}$$

Now solve the equation x=\frac{0±16}{4} when ± is plus. Divide 16 by 4.

$$x=4$$

Now solve the equation x=\frac{0±16}{4} when ± is minus. Divide -16 by 4.

$$x=-4$$

The equation is now solved.

$$x=4 x=-4$$