6=2(13+2f)f=

Solution for 6=2(13+2f)f= equation:

6=2(13+2f)f=

We move all terms to the left:

6-(2(13+2f)f)=0

We add all the numbers together, and all the variables

-(2(2f+13)f)+6=0


We calculate terms in parentheses: -(2(2f+13)f), so:

2(2f+13)f

We multiply parentheses

4f^2+26f

Back to the equation:

-(4f^2+26f)


We get rid of parentheses

-4f^2-26f+6=0

a = -4; b = -26; c = +6;
Δ = b2-4ac
Δ = -262-4·(-4)·6
Δ = 772
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:

$\sqrt{\Delta}=\sqrt{772}=\sqrt{4*193}=\sqrt{4}*\sqrt{193}=2\sqrt{193}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{193}}{2*-4}=\frac{26-2\sqrt{193}}{-8}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{193}}{2*-4}=\frac{26+2\sqrt{193}}{-8}
$