3^2+b^2=6^2

Solution for 3^2+b^2=6^2 equation:

3^2+b^2=6^2

We move all terms to the left:

3^2+b^2-(6^2)=0

We add all the numbers together, and all the variables

b^2-27=0

a = 1; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·1·(-27)
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{3}}{2*1}=\frac{0-6\sqrt{3}}{2}
=-\frac{6\sqrt{3}}{2}
=-3\sqrt{3}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{3}}{2*1}=\frac{0+6\sqrt{3}}{2}
=\frac{6\sqrt{3}}{2}
=3\sqrt{3}
$