3/4y-14+12-2/3y=40

Solution for 3/4y-14+12-2/3y=40 equation:

3/4y-14+12-2/3y=40

We move all terms to the left:

3/4y-14+12-2/3y-(40)=0


Domain of the equation: 4y!=0

y!=0/4

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

3/4y-2/3y-42=0

We calculate fractions


9y/12y^2+(-8y)/12y^2-42=0

We multiply all the terms by the denominator


9y+(-8y)-42*12y^2=0

Wy multiply elements

-504y^2+9y+(-8y)=0

We get rid of parentheses

-504y^2+9y-8y=0

We add all the numbers together, and all the variables

-504y^2+y=0

a = -504; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-504)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-504}=\frac{-2}{-1008}
=1/504
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-504}=\frac{0}{-1008}
=0
$