y+0.32y=7/5y

Solution for y+0.32y=7/5y equation:

y+0.32y=7/5y

We move all terms to the left:

y+0.32y-(7/5y)=0


Domain of the equation: 5y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

y+0.32y-(+7/5y)=0

We add all the numbers together, and all the variables

1.32y-(+7/5y)=0

We get rid of parentheses

1.32y-7/5y=0

We multiply all the terms by the denominator


(1.32y)*5y-7=0

We add all the numbers together, and all the variables

(+1.32y)*5y-7=0

We multiply parentheses

5y^2-7=0

a = 5; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·5·(-7)
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{35}}{2*5}=\frac{0-2\sqrt{35}}{10}
=-\frac{2\sqrt{35}}{10}
=-\frac{\sqrt{35}}{5}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{35}}{2*5}=\frac{0+2\sqrt{35}}{10}
=\frac{2\sqrt{35}}{10}
=\frac{\sqrt{35}}{5}
$