1.2x^2+.8x=0

Solution for 1.2x^2+.8x=0 equation:

1.2x^2+.8x=0

a = 1.2; b = .8; c = 0;
Δ = b2-4ac
Δ = .82-4·1.2·0
Δ = 0.64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(.8)-\sqrt{0.64}}{2*1.2}=\frac{-0.8-\sqrt{0.64}}{2.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.8)+\sqrt{0.64}}{2*1.2}=\frac{-0.8+\sqrt{0.64}}{2.4}
$