10^2+b^2=20^2

Solution for 10^2+b^2=20^2 equation:

10^2+b^2=20^2

We move all terms to the left:

10^2+b^2-(20^2)=0

We add all the numbers together, and all the variables

b^2-300=0

a = 1; b = 0; c = -300;
Δ = b2-4ac
Δ = 02-4·1·(-300)
Δ = 1200
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:

$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{3}}{2*1}=\frac{0-20\sqrt{3}}{2}
=-\frac{20\sqrt{3}}{2}
=-10\sqrt{3}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{3}}{2*1}=\frac{0+20\sqrt{3}}{2}
=\frac{20\sqrt{3}}{2}
=10\sqrt{3}
$