Find the Roots (Zeros) y=4-x^2

$y = 4 - x^{2}$

Set $4 - x^{2}$ equal to $0$ .

$4 - x^{2} = 0$

Solve for $x$ .

Subtract $4$ from both sides of the equation.

$- x^{2} = - 4$

Divide each term in $- x^{2} = - 4$ by $- 1$ and simplify.

Divide each term in $- x^{2} = - 4$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{- 4}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 4}{- 1}$

Simplify the right side.

Divide $- 4$ by $- 1$ .

$x^{2} = 4$

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

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

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

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

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.

$x = 2, - 2$