f(2)=5(2)+40

Solution for f(2)=5(2)+40 equation:

f(2)=5(2)+40

We move all terms to the left:

f(2)-(5(2)+40)=0

We add all the numbers together, and all the variables

f2-92=0

We add all the numbers together, and all the variables

f^2-92=0

a = 1; b = 0; c = -92;
Δ = b2-4ac
Δ = 02-4·1·(-92)
Δ = 368
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:

$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{23}}{2*1}=\frac{0-4\sqrt{23}}{2}
=-\frac{4\sqrt{23}}{2}
=-2\sqrt{23}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{23}}{2*1}=\frac{0+4\sqrt{23}}{2}
=\frac{4\sqrt{23}}{2}
=2\sqrt{23}
$