-0.5x^2+150x=0

Solution for -0.5x^2+150x=0 equation:

-0.5x^2+150x=0

a = -0.5; b = 150; c = 0;
Δ = b2-4ac
Δ = 1502-4·(-0.5)·0
Δ = 22500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{22500}=150$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-150}{2*-0.5}=\frac{-300}{-1}
=+300
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+150}{2*-0.5}=\frac{0}{-1}
=0
$