Solve for x x^2=4(4+21)

$x^{2} = 4 (4 + 21)$

Add $4$ and $21$ .

$x^{2} = 4 \cdot 25$

Multiply $4$ by $25$ .

$x^{2} = 100$

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

$x = \pm \sqrt{100}$

Simplify $\pm \sqrt{100}$ .

Rewrite $100$ as $10^{2}$ .

$x = \pm \sqrt{10^{2}}$

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

$x = \pm 10$

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.

$x = 10$

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

$x = - 10$

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

$x = 10, - 10$