(2/3)y+5=20-y

Solution for (2/3)y+5=20-y equation:

(2/3)y+5=20-y

We move all terms to the left:

(2/3)y+5-(20-y)=0


Domain of the equation: 3)y!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

(+2/3)y-(-1y+20)+5=0

We multiply parentheses

2y^2-(-1y+20)+5=0

We get rid of parentheses

2y^2+1y-20+5=0

We add all the numbers together, and all the variables

2y^2+y-15=0

a = 2; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·2·(-15)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*2}=\frac{-12}{4}
=-3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*2}=\frac{10}{4}
=2+1/2
$