y2+y/3=2

Solution for y2+y/3=2 equation:

y2+y/3=2

We move all terms to the left:

y2+y/3-(2)=0

We add all the numbers together, and all the variables

y^2+y/3-2=0

We multiply all the terms by the denominator


y^2*3+y-2*3=0

We add all the numbers together, and all the variables

y^2*3+y-6=0

Wy multiply elements

3y^2+y-6=0

a = 3; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·3·(-6)
Δ = 73
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{73}}{2*3}=\frac{-1-\sqrt{73}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{73}}{2*3}=\frac{-1+\sqrt{73}}{6}
$