1/4y+8=1/10y

Solution for 1/4y+8=1/10y equation:

1/4y+8=1/10y

We move all terms to the left:

1/4y+8-(1/10y)=0


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R



Domain of the equation: 10y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

1/4y-(+1/10y)+8=0

We get rid of parentheses

1/4y-1/10y+8=0

We calculate fractions


10y/40y^2+(-4y)/40y^2+8=0

We multiply all the terms by the denominator


10y+(-4y)+8*40y^2=0

Wy multiply elements

320y^2+10y+(-4y)=0

We get rid of parentheses

320y^2+10y-4y=0

We add all the numbers together, and all the variables

320y^2+6y=0

a = 320; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·320·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*320}=\frac{-12}{640}
=-3/160
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*320}=\frac{0}{640}
=0
$