F=3x+4/4x-5

Solution for F=3x+4/4x-5 equation:

=3F+4/4F-5

We move all terms to the left:

-(3F+4/4F-5)=0


Domain of the equation: 4F-5)!=0

F∈R


We get rid of parentheses

-3F-4/4F+5=0

We multiply all the terms by the denominator


-3F*4F+5*4F-4=0

Wy multiply elements

-12F^2+20F-4=0

a = -12; b = 20; c = -4;
Δ = b2-4ac
Δ = 202-4·(-12)·(-4)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{13}}{2*-12}=\frac{-20-4\sqrt{13}}{-24}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{13}}{2*-12}=\frac{-20+4\sqrt{13}}{-24}
$