169+b^2=324

Solution for 169+b^2=324 equation:

169+b^2=324

We move all terms to the left:

169+b^2-(324)=0

We add all the numbers together, and all the variables

b^2-155=0

a = 1; b = 0; c = -155;
Δ = b2-4ac
Δ = 02-4·1·(-155)
Δ = 620
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:

$\sqrt{\Delta}=\sqrt{620}=\sqrt{4*155}=\sqrt{4}*\sqrt{155}=2\sqrt{155}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{155}}{2*1}=\frac{0-2\sqrt{155}}{2}
=-\frac{2\sqrt{155}}{2}
=-\sqrt{155}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{155}}{2*1}=\frac{0+2\sqrt{155}}{2}
=\frac{2\sqrt{155}}{2}
=\sqrt{155}
$