5/4y=40,y

Solution for 5/4y=40,y equation:

5/4y=40.y

We move all terms to the left:

5/4y-(40.y)=0


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R


We add all the numbers together, and all the variables

5/4y-(+40.y)=0

We get rid of parentheses

5/4y-40.y=0

We multiply all the terms by the denominator


-(40.y)*4y+5=0

We add all the numbers together, and all the variables

-(+40.y)*4y+5=0

We multiply parentheses

-160y^2+5=0

a = -160; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-160)·5
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{2}}{2*-160}=\frac{0-40\sqrt{2}}{-320}
=-\frac{40\sqrt{2}}{-320}
=-\frac{\sqrt{2}}{-8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{2}}{2*-160}=\frac{0+40\sqrt{2}}{-320}
=\frac{40\sqrt{2}}{-320}
=\frac{\sqrt{2}}{-8}
$