5f(2f-4)=140

Solution for 5f(2f-4)=140 equation:

5f(2f-4)=140

We move all terms to the left:

5f(2f-4)-(140)=0

We multiply parentheses

10f^2-20f-140=0

a = 10; b = -20; c = -140;
Δ = b2-4ac
Δ = -202-4·10·(-140)
Δ = 6000
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:

$\sqrt{\Delta}=\sqrt{6000}=\sqrt{400*15}=\sqrt{400}*\sqrt{15}=20\sqrt{15}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20\sqrt{15}}{2*10}=\frac{20-20\sqrt{15}}{20}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20\sqrt{15}}{2*10}=\frac{20+20\sqrt{15}}{20}
$