f(2)=2(2)^2+4

Solution for f(2)=2(2)^2+4 equation:

f(2)=2(2)^2+4

We move all terms to the left:

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

We add all the numbers together, and all the variables

f^2-488=0

a = 1; b = 0; c = -488;
Δ = b2-4ac
Δ = 02-4·1·(-488)
Δ = 1952
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:

$\sqrt{\Delta}=\sqrt{1952}=\sqrt{16*122}=\sqrt{16}*\sqrt{122}=4\sqrt{122}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{122}}{2*1}=\frac{0-4\sqrt{122}}{2}
=-\frac{4\sqrt{122}}{2}
=-2\sqrt{122}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{122}}{2*1}=\frac{0+4\sqrt{122}}{2}
=\frac{4\sqrt{122}}{2}
=2\sqrt{122}
$