2/3-3/2y+1/3y+4=0

Solution for 2/3-3/2y+1/3y+4=0 equation:

2/3-3/2y+1/3y+4=0


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R



Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We calculate fractions


(-81y)/54y^2+2y/54y^2+4y/54y^2+4=0

We multiply all the terms by the denominator


(-81y)+2y+4y+4*54y^2=0

We add all the numbers together, and all the variables

6y+(-81y)+4*54y^2=0

Wy multiply elements

216y^2+6y+(-81y)=0

We get rid of parentheses

216y^2+6y-81y=0

We add all the numbers together, and all the variables

216y^2-75y=0

a = 216; b = -75; c = 0;
Δ = b2-4ac
Δ = -752-4·216·0
Δ = 5625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5625}=75$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-75)-75}{2*216}=\frac{0}{432}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-75)+75}{2*216}=\frac{150}{432}
=25/72
$