-6f+13=f2-11

Solution for -6f+13=f2-11 equation:

-6f+13=f2-11

We move all terms to the left:

-6f+13-(f2-11)=0

We add all the numbers together, and all the variables

-(+f^2-11)-6f+13=0

We get rid of parentheses

-f^2-6f+11+13=0

We add all the numbers together, and all the variables

-1f^2-6f+24=0

a = -1; b = -6; c = +24;
Δ = b2-4ac
Δ = -62-4·(-1)·24
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{33}}{2*-1}=\frac{6-2\sqrt{33}}{-2}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{33}}{2*-1}=\frac{6+2\sqrt{33}}{-2}
$