14^2+b^2=18^2

Solution for 14^2+b^2=18^2 equation:

14^2+b^2=18^2

We move all terms to the left:

14^2+b^2-(18^2)=0

We add all the numbers together, and all the variables

b^2-128=0

a = 1; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·1·(-128)
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{2}}{2*1}=\frac{0-16\sqrt{2}}{2}
=-\frac{16\sqrt{2}}{2}
=-8\sqrt{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{2}}{2*1}=\frac{0+16\sqrt{2}}{2}
=\frac{16\sqrt{2}}{2}
=8\sqrt{2}
$