11f-(2f+2);f=1/2

Solution for 11f-(2f+2);f=1/2 equation:

11f-(2f+2)f=1/2

We move all terms to the left:

11f-(2f+2)f-(1/2)=0

We add all the numbers together, and all the variables

11f-(2f+2)f-(+1/2)=0

We multiply parentheses

-2f^2+11f-2f-(+1/2)=0

We get rid of parentheses

-2f^2+11f-2f-1/2=0

We multiply all the terms by the denominator


-2f^2*2+11f*2-2f*2-1=0

Wy multiply elements

-4f^2+22f-4f-1=0

We add all the numbers together, and all the variables

-4f^2+18f-1=0

a = -4; b = 18; c = -1;
Δ = b2-4ac
Δ = 182-4·(-4)·(-1)
Δ = 308
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:

$\sqrt{\Delta}=\sqrt{308}=\sqrt{4*77}=\sqrt{4}*\sqrt{77}=2\sqrt{77}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{77}}{2*-4}=\frac{-18-2\sqrt{77}}{-8}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{77}}{2*-4}=\frac{-18+2\sqrt{77}}{-8}
$