-0.6x2+3.6x=0

Solution for -0.6x2+3.6x=0 equation:

-0.6x^2+3.6x=0

a = -0.6; b = 3.6; c = 0;
Δ = b2-4ac
Δ = 3.62-4·(-0.6)·0
Δ = 12.96
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{-(3.6)-\sqrt{12.96}}{2*-0.6}=\frac{-3.6-\sqrt{12.96}}{-1.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.6)+\sqrt{12.96}}{2*-0.6}=\frac{-3.6+\sqrt{12.96}}{-1.2}
$