.3y^2+3y+-4=-4

Solution for .3y^2+3y+-4=-4 equation:

.3y^2+3y+-4=-4

We move all terms to the left:

.3y^2+3y+-4-(-4)=0

We add all the numbers together, and all the variables

.3y^2+3y=0

a = .3; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·.3·0
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{-(3)-3}{2*.3}=\frac{-6}{0.6}
=-10
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*.3}=\frac{0}{0.6}
=0
$