-y^2+35.y-300=0

Solution for -y^2+35.y-300=0 equation:

-y^2+35.y-300=0

We add all the numbers together, and all the variables

-1y^2+35.y-300=0

a = -1; b = 35.; c = -300;
Δ = b2-4ac
Δ = 35.2-4·(-1)·(-300)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35.)-5}{2*-1}=\frac{-40}{-2}
=+20
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35.)+5}{2*-1}=\frac{-30}{-2}
=+15
$