-0.6x^2+36x=0

Solution for -0.6x^2+36x=0 equation:

-0.6x^2+36x=0

a = -0.6; b = 36; c = 0;
Δ = b2-4ac
Δ = 362-4·(-0.6)·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-36}{2*-0.6}=\frac{-72}{-1.2}
=+60
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+36}{2*-0.6}=\frac{0}{-1.2}
=0
$