b^2-96+18=0

Solution for b^2-96+18=0 equation:

b^2-96+18=0

We add all the numbers together, and all the variables

b^2-78=0

a = 1; b = 0; c = -78;
Δ = b2-4ac
Δ = 02-4·1·(-78)
Δ = 312
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:

$\sqrt{\Delta}=\sqrt{312}=\sqrt{4*78}=\sqrt{4}*\sqrt{78}=2\sqrt{78}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{78}}{2*1}=\frac{0-2\sqrt{78}}{2}
=-\frac{2\sqrt{78}}{2}
=-\sqrt{78}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{78}}{2*1}=\frac{0+2\sqrt{78}}{2}
=\frac{2\sqrt{78}}{2}
=\sqrt{78}
$