4b^2-4=96

Solution for 4b^2-4=96 equation:

4b^2-4=96

We move all terms to the left:

4b^2-4-(96)=0

We add all the numbers together, and all the variables

4b^2-100=0

a = 4; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·4·(-100)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*4}=\frac{-40}{8}
=-5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*4}=\frac{40}{8}
=5
$