y-(4-3y)/2y-(3+44)=1/5

Solution for y-(4-3y)/2y-(3+44)=1/5 equation:

y-(4-3y)/2y-(3+44)=1/5

We move all terms to the left:

y-(4-3y)/2y-(3+44)-(1/5)=0


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R


We add all the numbers together, and all the variables

y-(-3y+4)/2y-47-(+1/5)=0

We get rid of parentheses

y-(-3y+4)/2y-47-1/5=0

We calculate fractions


y+(15y-20)/10y+(-2y)/10y-47=0

We multiply all the terms by the denominator


y*10y+(15y-20)+(-2y)-47*10y=0

Wy multiply elements

10y^2+(15y-20)+(-2y)-470y=0

We get rid of parentheses

10y^2+15y-2y-470y-20=0

We add all the numbers together, and all the variables

10y^2-457y-20=0

a = 10; b = -457; c = -20;
Δ = b2-4ac
Δ = -4572-4·10·(-20)
Δ = 209649
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-457)-\sqrt{209649}}{2*10}=\frac{457-\sqrt{209649}}{20}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-457)+\sqrt{209649}}{2*10}=\frac{457+\sqrt{209649}}{20}
$