3(2^2x+5)=96

Solution for 3(2^2x+5)=96 equation:

3(2^2x+5)=96

We move all terms to the left:

3(2^2x+5)-(96)=0

We multiply parentheses

6x^2+15-96=0

We add all the numbers together, and all the variables

6x^2-81=0

a = 6; b = 0; c = -81;
Δ = b2-4ac
Δ = 02-4·6·(-81)
Δ = 1944
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1944}=\sqrt{324*6}=\sqrt{324}*\sqrt{6}=18\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{6}}{2*6}=\frac{0-18\sqrt{6}}{12}
=-\frac{18\sqrt{6}}{12}
=-\frac{3\sqrt{6}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{6}}{2*6}=\frac{0+18\sqrt{6}}{12}
=\frac{18\sqrt{6}}{12}
=\frac{3\sqrt{6}}{2}
$