4(3b^2-3)=288

Solution for 4(3b^2-3)=288 equation:

4(3b^2-3)=288

We move all terms to the left:

4(3b^2-3)-(288)=0

We multiply parentheses

12b^2-12-288=0

We add all the numbers together, and all the variables

12b^2-300=0

a = 12; b = 0; c = -300;
Δ = b2-4ac
Δ = 02-4·12·(-300)
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-120}{2*12}=\frac{-120}{24}
=-5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+120}{2*12}=\frac{120}{24}
=5
$