-2(2f-4)=-4(-f+2)f=

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

-2(2f-4)=-4(-f+2)f=

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-4f-(-4(-1f+2)f)+8=0


We calculate terms in parentheses: -(-4(-1f+2)f), so:

-4(-1f+2)f

We multiply parentheses

4f^2-8f

Back to the equation:

-(4f^2-8f)


We get rid of parentheses

-4f^2-4f+8f+8=0

We add all the numbers together, and all the variables

-4f^2+4f+8=0

a = -4; b = 4; c = +8;
Δ = b2-4ac
Δ = 42-4·(-4)·8
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*-4}=\frac{-16}{-8}
=+2
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*-4}=\frac{8}{-8}
=-1
$