-y^2+3y+13=0

Solution for -y^2+3y+13=0 equation:

-y^2+3y+13=0

We add all the numbers together, and all the variables

-1y^2+3y+13=0

a = -1; b = 3; c = +13;
Δ = b2-4ac
Δ = 32-4·(-1)·13
Δ = 61
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{61}}{2*-1}=\frac{-3-\sqrt{61}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{61}}{2*-1}=\frac{-3+\sqrt{61}}{-2}
$