f(2)=50(2)+100

Solution for f(2)=50(2)+100 equation:

f(2)=50(2)+100

We move all terms to the left:

f(2)-(50(2)+100)=0

We add all the numbers together, and all the variables

f2-602=0

We add all the numbers together, and all the variables

f^2-602=0

a = 1; b = 0; c = -602;
Δ = b2-4ac
Δ = 02-4·1·(-602)
Δ = 2408
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:

$\sqrt{\Delta}=\sqrt{2408}=\sqrt{4*602}=\sqrt{4}*\sqrt{602}=2\sqrt{602}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{602}}{2*1}=\frac{0-2\sqrt{602}}{2}
=-\frac{2\sqrt{602}}{2}
=-\sqrt{602}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{602}}{2*1}=\frac{0+2\sqrt{602}}{2}
=\frac{2\sqrt{602}}{2}
=\sqrt{602}
$