18^2+b^2=21^2

Solution for 18^2+b^2=21^2 equation:

18^2+b^2=21^2

We move all terms to the left:

18^2+b^2-(21^2)=0

We add all the numbers together, and all the variables

b^2-117=0

a = 1; b = 0; c = -117;
Δ = b2-4ac
Δ = 02-4·1·(-117)
Δ = 468
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:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{13}}{2*1}=\frac{0-6\sqrt{13}}{2}
=-\frac{6\sqrt{13}}{2}
=-3\sqrt{13}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{13}}{2*1}=\frac{0+6\sqrt{13}}{2}
=\frac{6\sqrt{13}}{2}
=3\sqrt{13}
$