3x+0.4x^2=0

Solution for 3x+0.4x^2=0 equation:

3x+0.4x^2=0

a = 0.4; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·0.4·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*0.4}=\frac{-6}{0.8}
=-7+0.4/0.8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*0.4}=\frac{0}{0.8}
=0
$