y2+4y-396=0

Solution for y2+4y-396=0 equation:

y2+4y-396=0

We add all the numbers together, and all the variables

y^2+4y-396=0

a = 1; b = 4; c = -396;
Δ = b2-4ac
Δ = 42-4·1·(-396)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-40}{2*1}=\frac{-44}{2}
=-22
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+40}{2*1}=\frac{36}{2}
=18
$