f(2)=2(2)^2-3(2)+7

Solution for f(2)=2(2)^2-3(2)+7 equation:

f(2)=2(2)^2-3(2)+7

We move all terms to the left:

f(2)-(2(2)^2-3(2)+7)=0

We add all the numbers together, and all the variables

f^2-459=0

a = 1; b = 0; c = -459;
Δ = b2-4ac
Δ = 02-4·1·(-459)
Δ = 1836
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:

$\sqrt{\Delta}=\sqrt{1836}=\sqrt{36*51}=\sqrt{36}*\sqrt{51}=6\sqrt{51}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{51}}{2*1}=\frac{0-6\sqrt{51}}{2}
=-\frac{6\sqrt{51}}{2}
=-3\sqrt{51}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{51}}{2*1}=\frac{0+6\sqrt{51}}{2}
=\frac{6\sqrt{51}}{2}
=3\sqrt{51}
$