-y^2+13y=40

Solution for -y^2+13y=40 equation:

-y^2+13y=40

We move all terms to the left:

-y^2+13y-(40)=0

We add all the numbers together, and all the variables

-1y^2+13y-40=0

a = -1; b = 13; c = -40;
Δ = b2-4ac
Δ = 132-4·(-1)·(-40)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3}{2*-1}=\frac{-16}{-2}
=+8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3}{2*-1}=\frac{-10}{-2}
=+5
$