7f(f-5)=28

Solution for 7f(f-5)=28 equation:

7f(f-5)=28

We move all terms to the left:

7f(f-5)-(28)=0

We multiply parentheses

7f^2-35f-28=0

a = 7; b = -35; c = -28;
Δ = b2-4ac
Δ = -352-4·7·(-28)
Δ = 2009
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:

$\sqrt{\Delta}=\sqrt{2009}=\sqrt{49*41}=\sqrt{49}*\sqrt{41}=7\sqrt{41}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-7\sqrt{41}}{2*7}=\frac{35-7\sqrt{41}}{14}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+7\sqrt{41}}{2*7}=\frac{35+7\sqrt{41}}{14}
$