64+b^2=17^2

Solution for 64+b^2=17^2 equation:

64+b^2=17^2

We move all terms to the left:

64+b^2-(17^2)=0

We add all the numbers together, and all the variables

b^2-225=0

a = 1; b = 0; c = -225;
Δ = b2-4ac
Δ = 02-4·1·(-225)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30}{2*1}=\frac{-30}{2}
=-15
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30}{2*1}=\frac{30}{2}
=15
$