F(x)=3x^+6x-3/x+2

Solution for F(x)=3x^+6x-3/x+2 equation:

(F)=3F^+6F-3/F+2

We move all terms to the left:

(F)-(3F^+6F-3/F+2)=0


Domain of the equation: F+2)!=0

F∈R


We get rid of parentheses

F-3F^-6F+3/F-2=0

We multiply all the terms by the denominator


F*F-3F^*F-6F*F-2*F+3=0

We add all the numbers together, and all the variables

-2F+F*F-3F^*F-6F*F+3=0

Wy multiply elements

F^2-3F^2-6F^2-2F+3=0

We add all the numbers together, and all the variables

-8F^2-2F+3=0

a = -8; b = -2; c = +3;
Δ = b2-4ac
Δ = -22-4·(-8)·3
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{-(-2)-10}{2*-8}=\frac{-8}{-16}
=1/2
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*-8}=\frac{12}{-16}
=-3/4
$