4/2y-5-3/y+3=0

Solution for 4/2y-5-3/y+3=0 equation:

4/2y-5-3/y+3=0


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R



Domain of the equation: y!=0

y∈R


We add all the numbers together, and all the variables

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

We calculate fractions


4y/2y^2+(-6y)/2y^2-2=0

We multiply all the terms by the denominator


4y+(-6y)-2*2y^2=0

Wy multiply elements

-4y^2+4y+(-6y)=0

We get rid of parentheses

-4y^2+4y-6y=0

We add all the numbers together, and all the variables

-4y^2-2y=0

a = -4; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-4)·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*-4}=\frac{0}{-8}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-4}=\frac{4}{-8}
=-1/2
$