#### Answer

$x=\pm\sqrt{12}$

#### Work Step by Step

$f(x)=\frac{12-x^2}{8}$
Since we are solving for all the real values of $x$ when $f(x) = 0$, set the function $f(x)=\frac{12-x^2}{8}$ equal to zero and we get the following equation:
$\frac{12-x^2}{8}=0$
First multipy both sides by 8 to simplify the denominator.
$\frac{12-x^2}{8}(8)=0(8)$
$12-x^2=0$
Now subtract 12 from both sides of the equation so that all of the constants are on the right side of the equation and all of the $x$ terms are on the left side.
$12-x^2-12=0-12$
$-x^2=-12$
Multiply both sides by -1 to simply the signs in the equation.
$x^2=12$
Once all of the constants are on the right and all the $x$ terms are on the left, solve for $x$ by taking the square root of both sides.
$\sqrt{x^2}=\sqrt{12}$
Note: Remember taking the square root of both sides of the equation results in both a positive and negative answer. This is why the $\pm$ appears in the line below.
$x=\pm\sqrt{12}$