3/2y+12=y-20

Solution for 3/2y+12=y-20 equation:

3/2y+12=y-20

We move all terms to the left:

3/2y+12-(y-20)=0


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R


We get rid of parentheses

3/2y-y+20+12=0

We multiply all the terms by the denominator


-y*2y+20*2y+12*2y+3=0

Wy multiply elements

-2y^2+40y+24y+3=0

We add all the numbers together, and all the variables

-2y^2+64y+3=0

a = -2; b = 64; c = +3;
Δ = b2-4ac
Δ = 642-4·(-2)·3
Δ = 4120
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:

$\sqrt{\Delta}=\sqrt{4120}=\sqrt{4*1030}=\sqrt{4}*\sqrt{1030}=2\sqrt{1030}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-2\sqrt{1030}}{2*-2}=\frac{-64-2\sqrt{1030}}{-4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+2\sqrt{1030}}{2*-2}=\frac{-64+2\sqrt{1030}}{-4}
$