-0.3x^2+75x-2000=0

Solution for -0.3x^2+75x-2000=0 equation:

-0.3x^2+75x-2000=0

a = -0.3; b = 75; c = -2000;
Δ = b2-4ac
Δ = 752-4·(-0.3)·(-2000)
Δ = 3225
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:

$\sqrt{\Delta}=\sqrt{3225}=\sqrt{25*129}=\sqrt{25}*\sqrt{129}=5\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(75)-5\sqrt{129}}{2*-0.3}=\frac{-75-5\sqrt{129}}{-0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(75)+5\sqrt{129}}{2*-0.3}=\frac{-75+5\sqrt{129}}{-0.6}
$