f(2)=3(4)^2-4

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

f(2)=3(4)^2-4

We move all terms to the left:

f(2)-(3(4)^2-4)=0

We add all the numbers together, and all the variables

f^2-1152=0

a = 1; b = 0; c = -1152;
Δ = b2-4ac
Δ = 02-4·1·(-1152)
Δ = 4608
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4608}=\sqrt{2304*2}=\sqrt{2304}*\sqrt{2}=48\sqrt{2}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{2}}{2*1}=\frac{0-48\sqrt{2}}{2}
=-\frac{48\sqrt{2}}{2}
=-24\sqrt{2}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{2}}{2*1}=\frac{0+48\sqrt{2}}{2}
=\frac{48\sqrt{2}}{2}
=24\sqrt{2}
$