2/3y+5=3y+19

Solution for 2/3y+5=3y+19 equation:

2/3y+5=3y+19

We move all terms to the left:

2/3y+5-(3y+19)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We get rid of parentheses

2/3y-3y-19+5=0

We multiply all the terms by the denominator


-3y*3y-19*3y+5*3y+2=0

Wy multiply elements

-9y^2-57y+15y+2=0

We add all the numbers together, and all the variables

-9y^2-42y+2=0

a = -9; b = -42; c = +2;
Δ = b2-4ac
Δ = -422-4·(-9)·2
Δ = 1836
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:

$\sqrt{\Delta}=\sqrt{1836}=\sqrt{36*51}=\sqrt{36}*\sqrt{51}=6\sqrt{51}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6\sqrt{51}}{2*-9}=\frac{42-6\sqrt{51}}{-18}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6\sqrt{51}}{2*-9}=\frac{42+6\sqrt{51}}{-18}
$