3(y-2)/4+4(2y-3)/5y=6

Solution for 3(y-2)/4+4(2y-3)/5y=6 equation:

3(y-2)/4+4(2y-3)/5y=6

We move all terms to the left:

3(y-2)/4+4(2y-3)/5y-(6)=0


Domain of the equation: 5y!=0

y!=0/5

y!=0

y∈R


We calculate fractions


(15y^2-30y)/20y+(32y-48)/20y-6=0

We multiply all the terms by the denominator


(15y^2-30y)+(32y-48)-6*20y=0

Wy multiply elements

(15y^2-30y)+(32y-48)-120y=0

We get rid of parentheses

15y^2-30y+32y-120y-48=0

We add all the numbers together, and all the variables

15y^2-118y-48=0

a = 15; b = -118; c = -48;
Δ = b2-4ac
Δ = -1182-4·15·(-48)
Δ = 16804
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{16804}=\sqrt{4*4201}=\sqrt{4}*\sqrt{4201}=2\sqrt{4201}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-118)-2\sqrt{4201}}{2*15}=\frac{118-2\sqrt{4201}}{30}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-118)+2\sqrt{4201}}{2*15}=\frac{118+2\sqrt{4201}}{30}
$