(40y/2y)+12=32y=

Solution for (40y/2y)+12=32y= equation:

(40y/2y)+12=32y=

We move all terms to the left:

(40y/2y)+12-(32y)=0


Domain of the equation: 2y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

(+40y/2y)-32y+12=0

We add all the numbers together, and all the variables

-32y+(+40y/2y)+12=0

We get rid of parentheses

-32y+40y/2y+12=0

We multiply all the terms by the denominator


-32y*2y+40y+12*2y=0

We add all the numbers together, and all the variables

40y-32y*2y+12*2y=0

Wy multiply elements

-64y^2+40y+24y=0

We add all the numbers together, and all the variables

-64y^2+64y=0

a = -64; b = 64; c = 0;
Δ = b2-4ac
Δ = 642-4·(-64)·0
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-64}{2*-64}=\frac{-128}{-128}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+64}{2*-64}=\frac{0}{-128}
=0
$