f(2)=3(2+4)

Solution for f(2)=3(2+4) equation:

f(2)=3(2+4)

We move all terms to the left:

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

We add all the numbers together, and all the variables

f2-(36)=0

We add all the numbers together, and all the variables

f^2-36=0

a = 1; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·1·(-36)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*1}=\frac{-12}{2}
=-6
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*1}=\frac{12}{2}
=6
$