12^2+b^2=16^2

Solution for 12^2+b^2=16^2 equation:

12^2+b^2=16^2

We move all terms to the left:

12^2+b^2-(16^2)=0

We add all the numbers together, and all the variables

b^2-112=0

a = 1; b = 0; c = -112;
Δ = b2-4ac
Δ = 02-4·1·(-112)
Δ = 448
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{7}}{2*1}=\frac{0-8\sqrt{7}}{2}
=-\frac{8\sqrt{7}}{2}
=-4\sqrt{7}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{7}}{2*1}=\frac{0+8\sqrt{7}}{2}
=\frac{8\sqrt{7}}{2}
=4\sqrt{7}
$