180-y=(2/3y)+10

Solution for 180-y=(2/3y)+10 equation:

180-y=(2/3y)+10

We move all terms to the left:

180-y-((2/3y)+10)=0


Domain of the equation: 3y)+10)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

-y-((+2/3y)+10)+180=0

We add all the numbers together, and all the variables

-1y-((+2/3y)+10)+180=0

We multiply all the terms by the denominator


-1y*3y)+10)-((+180*3y)+10)+2=0

Wy multiply elements

-3y^2+540y=0

a = -3; b = 540; c = 0;
Δ = b2-4ac
Δ = 5402-4·(-3)·0
Δ = 291600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{291600}=540$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(540)-540}{2*-3}=\frac{-1080}{-6}
=+180
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(540)+540}{2*-3}=\frac{0}{-6}
=0
$