(3y-6)+(7y-2)+17/3y=180

Solution for (3y-6)+(7y-2)+17/3y=180 equation:

(3y-6)+(7y-2)+17/3y=180

We move all terms to the left:

(3y-6)+(7y-2)+17/3y-(180)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We get rid of parentheses

3y+7y+17/3y-6-2-180=0

We multiply all the terms by the denominator


3y*3y+7y*3y-6*3y-2*3y-180*3y+17=0

Wy multiply elements

9y^2+21y^2-18y-6y-540y+17=0

We add all the numbers together, and all the variables

30y^2-564y+17=0

a = 30; b = -564; c = +17;
Δ = b2-4ac
Δ = -5642-4·30·17
Δ = 316056
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:

$\sqrt{\Delta}=\sqrt{316056}=\sqrt{4*79014}=\sqrt{4}*\sqrt{79014}=2\sqrt{79014}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-564)-2\sqrt{79014}}{2*30}=\frac{564-2\sqrt{79014}}{60}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-564)+2\sqrt{79014}}{2*30}=\frac{564+2\sqrt{79014}}{60}
$