16=4+x^2

Solution for 16=4+x^2 equation:

16=4+x^2

We move all terms to the left:

16-(4+x^2)=0

We get rid of parentheses

-x^2-4+16=0

We add all the numbers together, and all the variables

-1x^2+12=0

a = -1; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-1)·12
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*-1}=\frac{0-4\sqrt{3}}{-2}
=-\frac{4\sqrt{3}}{-2}
=-\frac{2\sqrt{3}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*-1}=\frac{0+4\sqrt{3}}{-2}
=\frac{4\sqrt{3}}{-2}
=\frac{2\sqrt{3}}{-1}
$