-2000+95x-0.2x^2=0

Solution for -2000+95x-0.2x^2=0 equation:

-2000+95x-0.2x^2=0

a = -0.2; b = 95; c = -2000;
Δ = b2-4ac
Δ = 952-4·(-0.2)·(-2000)
Δ = 7425
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:

$\sqrt{\Delta}=\sqrt{7425}=\sqrt{225*33}=\sqrt{225}*\sqrt{33}=15\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(95)-15\sqrt{33}}{2*-0.2}=\frac{-95-15\sqrt{33}}{-0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(95)+15\sqrt{33}}{2*-0.2}=\frac{-95+15\sqrt{33}}{-0.4}
$