2/3y+y-4=3-

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

2/3y+y-4=3-

We move all terms to the left:

2/3y+y-4-(3-)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

y+2/3y-4=0

We multiply all the terms by the denominator


y*3y-4*3y+2=0

Wy multiply elements

3y^2-12y+2=0

a = 3; b = -12; c = +2;
Δ = b2-4ac
Δ = -122-4·3·2
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{30}}{2*3}=\frac{12-2\sqrt{30}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{30}}{2*3}=\frac{12+2\sqrt{30}}{6}
$