-0,3x^2+2,4x=0

Solution for -0,3x^2+2,4x=0 equation:

-0.3x^2+2.4x=0

a = -0.3; b = 2.4; c = 0;
Δ = b2-4ac
Δ = 2.42-4·(-0.3)·0
Δ = 5.76
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{-(2.4)-\sqrt{5.76}}{2*-0.3}=\frac{-2.4-\sqrt{5.76}}{-0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.4)+\sqrt{5.76}}{2*-0.3}=\frac{-2.4+\sqrt{5.76}}{-0.6}
$