-0.2x^2+100x-1500=0

Solution for -0.2x^2+100x-1500=0 equation:

-0.2x^2+100x-1500=0

a = -0.2; b = 100; c = -1500;
Δ = b2-4ac
Δ = 1002-4·(-0.2)·(-1500)
Δ = 8800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8800}=\sqrt{400*22}=\sqrt{400}*\sqrt{22}=20\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-20\sqrt{22}}{2*-0.2}=\frac{-100-20\sqrt{22}}{-0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+20\sqrt{22}}{2*-0.2}=\frac{-100+20\sqrt{22}}{-0.4}
$