-31/2y+5+62/3y+4=12

Solution for -31/2y+5+62/3y+4=12 equation:

-31/2y+5+62/3y+4=12

We move all terms to the left:

-31/2y+5+62/3y+4-(12)=0


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R



Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We add all the numbers together, and all the variables

-31/2y+62/3y-3=0

We calculate fractions


(-93y)/6y^2+124y/6y^2-3=0

We multiply all the terms by the denominator


(-93y)+124y-3*6y^2=0

We add all the numbers together, and all the variables

124y+(-93y)-3*6y^2=0

Wy multiply elements

-18y^2+124y+(-93y)=0

We get rid of parentheses

-18y^2+124y-93y=0

We add all the numbers together, and all the variables

-18y^2+31y=0

a = -18; b = 31; c = 0;
Δ = b2-4ac
Δ = 312-4·(-18)·0
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-31}{2*-18}=\frac{-62}{-36}
=1+13/18
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+31}{2*-18}=\frac{0}{-36}
=0
$