-14f+f^2-18=-8f-2

Solution for -14f+f^2-18=-8f-2 equation:

-14f+f^2-18=-8f-2

We move all terms to the left:

-14f+f^2-18-(-8f-2)=0

We get rid of parentheses

f^2-14f+8f+2-18=0

We add all the numbers together, and all the variables

f^2-6f-16=0

a = 1; b = -6; c = -16;
Δ = b2-4ac
Δ = -62-4·1·(-16)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-10}{2*1}=\frac{-4}{2}
=-2
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+10}{2*1}=\frac{16}{2}
=8
$