3y-y^2+3=0

Solution for 3y-y^2+3=0 equation:

3y-y^2+3=0

We add all the numbers together, and all the variables

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

a = -1; b = 3; c = +3;
Δ = b2-4ac
Δ = 32-4·(-1)·3
Δ = 21
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{21}}{2*-1}=\frac{-3-\sqrt{21}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{21}}{2*-1}=\frac{-3+\sqrt{21}}{-2}
$