.31=3y+y^2

Solution for .31=3y+y^2 equation:

.31=3y+y^2

We move all terms to the left:

.31-(3y+y^2)=0

We add all the numbers together, and all the variables

-(3y+y^2)+0.31=0

We get rid of parentheses

-y^2-3y+0.31=0

We add all the numbers together, and all the variables

-1y^2-3y+0.31=0

a = -1; b = -3; c = +0.31;
Δ = b2-4ac
Δ = -32-4·(-1)·0.31
Δ = 10.24
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{-(-3)-\sqrt{10.24}}{2*-1}=\frac{3-\sqrt{10.24}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{10.24}}{2*-1}=\frac{3+\sqrt{10.24}}{-2}
$