a^2+16=152

Solution for a^2+16=152 equation:

a^2+16=152

We move all terms to the left:

a^2+16-(152)=0

We add all the numbers together, and all the variables

a^2-136=0

a = 1; b = 0; c = -136;
Δ = b2-4ac
Δ = 02-4·1·(-136)
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{34}}{2*1}=\frac{0-4\sqrt{34}}{2}
=-\frac{4\sqrt{34}}{2}
=-2\sqrt{34}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{34}}{2*1}=\frac{0+4\sqrt{34}}{2}
=\frac{4\sqrt{34}}{2}
=2\sqrt{34}
$