y=3,-y^2

Solution for y=3,-y^2 equation:

y=3.-y^2

We move all terms to the left:

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

We get rid of parentheses

y^2+y-3.=0

We add all the numbers together, and all the variables

y^2+y-3=0

a = 1; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·1·(-3)
Δ = 13
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{13}}{2*1}=\frac{-1-\sqrt{13}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{13}}{2*1}=\frac{-1+\sqrt{13}}{2}
$