121=36+b^2

Solution for 121=36+b^2 equation:

121=36+b^2

We move all terms to the left:

121-(36+b^2)=0

We get rid of parentheses

-b^2-36+121=0

We add all the numbers together, and all the variables

-1b^2+85=0

a = -1; b = 0; c = +85;
Δ = b2-4ac
Δ = 02-4·(-1)·85
Δ = 340
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:

$\sqrt{\Delta}=\sqrt{340}=\sqrt{4*85}=\sqrt{4}*\sqrt{85}=2\sqrt{85}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{85}}{2*-1}=\frac{0-2\sqrt{85}}{-2}
=-\frac{2\sqrt{85}}{-2}
=-\frac{\sqrt{85}}{-1}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{85}}{2*-1}=\frac{0+2\sqrt{85}}{-2}
=\frac{2\sqrt{85}}{-2}
=\frac{\sqrt{85}}{-1}
$