2f(3-f)=4

Solution for 2f(3-f)=4 equation:

2f(3-f)=4

We move all terms to the left:

2f(3-f)-(4)=0

We add all the numbers together, and all the variables

2f(-1f+3)-4=0

We multiply parentheses

-2f^2+6f-4=0

a = -2; b = 6; c = -4;
Δ = b2-4ac
Δ = 62-4·(-2)·(-4)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*-2}=\frac{-8}{-4}
=+2
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*-2}=\frac{-4}{-4}
=1
$