12^2+b^2=27^2

Solution for 12^2+b^2=27^2 equation:

12^2+b^2=27^2

We move all terms to the left:

12^2+b^2-(27^2)=0

We add all the numbers together, and all the variables

b^2-585=0

a = 1; b = 0; c = -585;
Δ = b2-4ac
Δ = 02-4·1·(-585)
Δ = 2340
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:

$\sqrt{\Delta}=\sqrt{2340}=\sqrt{36*65}=\sqrt{36}*\sqrt{65}=6\sqrt{65}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{65}}{2*1}=\frac{0-6\sqrt{65}}{2}
=-\frac{6\sqrt{65}}{2}
=-3\sqrt{65}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{65}}{2*1}=\frac{0+6\sqrt{65}}{2}
=\frac{6\sqrt{65}}{2}
=3\sqrt{65}
$