f(2)=2(2)^2

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

f(2)=2(2)^2

We move all terms to the left:

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

determiningTheFunctionDomain
f2-22^2=0

We add all the numbers together, and all the variables

f^2-484=0

a = 1; b = 0; c = -484;
Δ = b2-4ac
Δ = 02-4·1·(-484)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-44}{2*1}=\frac{-44}{2}
=-22
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+44}{2*1}=\frac{44}{2}
=22
$