-0.3x^2+1,5x=0

Solution for -0.3x^2+1,5x=0 equation:

-0.3x^2+1.5x=0

a = -0.3; b = 1.5; c = 0;
Δ = b2-4ac
Δ = 1.52-4·(-0.3)·0
Δ = 2.25
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{-(1.5)-\sqrt{2.25}}{2*-0.3}=\frac{-1.5-\sqrt{2.25}}{-0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.5)+\sqrt{2.25}}{2*-0.3}=\frac{-1.5+\sqrt{2.25}}{-0.6}
$