(2/3)y+y=35

Solution for (2/3)y+y=35 equation:

(2/3)y+y=35

We move all terms to the left:

(2/3)y+y-(35)=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+y-35=0

We add all the numbers together, and all the variables

y+(+2/3)y-35=0

We multiply parentheses

2y^2+y-35=0

a = 2; b = 1; c = -35;
Δ = b2-4ac
Δ = 12-4·2·(-35)
Δ = 281
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{281}}{2*2}=\frac{-1-\sqrt{281}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{281}}{2*2}=\frac{-1+\sqrt{281}}{4}
$