16^2+b^2=65^2

Solution for 16^2+b^2=65^2 equation:

16^2+b^2=65^2

We move all terms to the left:

16^2+b^2-(65^2)=0

We add all the numbers together, and all the variables

b^2-3969=0

a = 1; b = 0; c = -3969;
Δ = b2-4ac
Δ = 02-4·1·(-3969)
Δ = 15876
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{15876}=126$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-126}{2*1}=\frac{-126}{2}
=-63
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+126}{2*1}=\frac{126}{2}
=63
$