-y^2+4y+35=0

Solution for -y^2+4y+35=0 equation:

-y^2+4y+35=0

We add all the numbers together, and all the variables

-1y^2+4y+35=0

a = -1; b = 4; c = +35;
Δ = b2-4ac
Δ = 42-4·(-1)·35
Δ = 156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{39}}{2*-1}=\frac{-4-2\sqrt{39}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{39}}{2*-1}=\frac{-4+2\sqrt{39}}{-2}
$