21^2+b^2=27^2

Solution for 21^2+b^2=27^2 equation:

21^2+b^2=27^2

We move all terms to the left:

21^2+b^2-(27^2)=0

We add all the numbers together, and all the variables

b^2-288=0

a = 1; b = 0; c = -288;
Δ = b2-4ac
Δ = 02-4·1·(-288)
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{2}}{2*1}=\frac{0-24\sqrt{2}}{2}
=-\frac{24\sqrt{2}}{2}
=-12\sqrt{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{2}}{2*1}=\frac{0+24\sqrt{2}}{2}
=\frac{24\sqrt{2}}{2}
=12\sqrt{2}
$