-y2+3y+1=0

Solution for -y2+3y+1=0 equation:

-y2+3y+1=0

We add all the numbers together, and all the variables

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

a = -1; b = 3; c = +1;
Δ = b2-4ac
Δ = 32-4·(-1)·1
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{-(3)-\sqrt{13}}{2*-1}=\frac{-3-\sqrt{13}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{13}}{2*-1}=\frac{-3+\sqrt{13}}{-2}
$