2y+(7y-1/2)*2=2(1-5/2y)

Solution for 2y+(7y-1/2)*2=2(1-5/2y) equation:

2y+(7y-1/2)*2=2(1-5/2y)

We move all terms to the left:

2y+(7y-1/2)*2-(2(1-5/2y))=0


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

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

2y+(+7y-1/2)*2-(2(-5/2y+1))=0

We multiply parentheses

2y+14y-(2(-5/2y+1))-1/2*2=0

We calculate fractions


2y+14y+()/2y+(-2y)/2y=0

We add all the numbers together, and all the variables

16y+()/2y+(-2y)/2y=0

We multiply all the terms by the denominator


16y*2y+(-2y)+()=0

We add all the numbers together, and all the variables

16y*2y+(-2y)=0

Wy multiply elements

32y^2+(-2y)=0

We get rid of parentheses

32y^2-2y=0

a = 32; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·32·0
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*32}=\frac{0}{64}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*32}=\frac{4}{64}
=1/16
$