4a+16=a^2+2

Solution for 4a+16=a^2+2 equation:

4a+16=a^2+2

We move all terms to the left:

4a+16-(a^2+2)=0

We get rid of parentheses

-a^2+4a-2+16=0

We add all the numbers together, and all the variables

-1a^2+4a+14=0

a = -1; b = 4; c = +14;
Δ = b2-4ac
Δ = 42-4·(-1)·14
Δ = 72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6\sqrt{2}}{2*-1}=\frac{-4-6\sqrt{2}}{-2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6\sqrt{2}}{2*-1}=\frac{-4+6\sqrt{2}}{-2}
$