Solve for x -2x^4+32=0

$- 2 x^{4} + 32 = 0$

Subtract $32$ from both sides of the equation.

$- 2 x^{4} = - 32$

Divide each term in $- 2 x^{4} = - 32$ by $- 2$ and simplify.

Divide each term in $- 2 x^{4} = - 32$ by $- 2$ .

$\frac{- 2 x^{4}}{- 2} = \frac{- 32}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x^{4}}{- 2} = \frac{- 32}{- 2}$

Divide $x^{4}$ by $1$ .

$x^{4} = \frac{- 32}{- 2}$

Simplify the right side.

Divide $- 32$ by $- 2$ .

$x^{4} = 16$

Take the 4th root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt[4]{16}$

Simplify $\pm \sqrt[4]{16}$ .

Rewrite $16$ as $2^{4}$ .

$x = \pm \sqrt[4]{2^{4}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$