b^2-18=16^2

Solution for b^2-18=16^2 equation:

b^2-18=16^2

We move all terms to the left:

b^2-18-(16^2)=0

We add all the numbers together, and all the variables

b^2-274=0

a = 1; b = 0; c = -274;
Δ = b2-4ac
Δ = 02-4·1·(-274)
Δ = 1096
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:

$\sqrt{\Delta}=\sqrt{1096}=\sqrt{4*274}=\sqrt{4}*\sqrt{274}=2\sqrt{274}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{274}}{2*1}=\frac{0-2\sqrt{274}}{2}
=-\frac{2\sqrt{274}}{2}
=-\sqrt{274}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{274}}{2*1}=\frac{0+2\sqrt{274}}{2}
=\frac{2\sqrt{274}}{2}
=\sqrt{274}
$