-0.2x^2+1.3x=0

Solution for -0.2x^2+1.3x=0 equation:

-0.2x^2+1.3x=0

a = -0.2; b = 1.3; c = 0;
Δ = b2-4ac
Δ = 1.32-4·(-0.2)·0
Δ = 1.69
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{-(1.3)-\sqrt{1.69}}{2*-0.2}=\frac{-1.3-\sqrt{1.69}}{-0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.3)+\sqrt{1.69}}{2*-0.2}=\frac{-1.3+\sqrt{1.69}}{-0.4}
$