F(x)=2x-3/3x+4

Solution for F(x)=2x-3/3x+4 equation:

(F)=2F-3/3F+4

We move all terms to the left:

(F)-(2F-3/3F+4)=0


Domain of the equation: 3F+4)!=0

F∈R


We get rid of parentheses

F-2F+3/3F-4=0

We multiply all the terms by the denominator


F*3F-2F*3F-4*3F+3=0

Wy multiply elements

3F^2-6F^2-12F+3=0

We add all the numbers together, and all the variables

-3F^2-12F+3=0

a = -3; b = -12; c = +3;
Δ = b2-4ac
Δ = -122-4·(-3)·3
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{5}}{2*-3}=\frac{12-6\sqrt{5}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{5}}{2*-3}=\frac{12+6\sqrt{5}}{-6}
$