1/4y-3=1/2y+12

Solution for 1/4y-3=1/2y+12 equation:

1/4y-3=1/2y+12

We move all terms to the left:

1/4y-3-(1/2y+12)=0


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R



Domain of the equation: 2y+12)!=0

y∈R


We get rid of parentheses

1/4y-1/2y-12-3=0

We calculate fractions


2y/8y^2+(-4y)/8y^2-12-3=0

We add all the numbers together, and all the variables

2y/8y^2+(-4y)/8y^2-15=0

We multiply all the terms by the denominator


2y+(-4y)-15*8y^2=0

Wy multiply elements

-120y^2+2y+(-4y)=0

We get rid of parentheses

-120y^2+2y-4y=0

We add all the numbers together, and all the variables

-120y^2-2y=0

a = -120; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-120)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-120}=\frac{0}{-240}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-120}=\frac{4}{-240}
=-1/60
$