f(2)=4(2)^2+5

Solution for f(2)=4(2)^2+5 equation:

f(2)=4(2)^2+5

We move all terms to the left:

f(2)-(4(2)^2+5)=0

We add all the numbers together, and all the variables

f^2-1769=0

a = 1; b = 0; c = -1769;
Δ = b2-4ac
Δ = 02-4·1·(-1769)
Δ = 7076
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:

$\sqrt{\Delta}=\sqrt{7076}=\sqrt{4*1769}=\sqrt{4}*\sqrt{1769}=2\sqrt{1769}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{1769}}{2*1}=\frac{0-2\sqrt{1769}}{2}
=-\frac{2\sqrt{1769}}{2}
=-\sqrt{1769}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{1769}}{2*1}=\frac{0+2\sqrt{1769}}{2}
=\frac{2\sqrt{1769}}{2}
=\sqrt{1769}
$