-0.2x2+90x-2000=0

Solution for -0.2x2+90x-2000=0 equation:

-0.2x^2+90x-2000=0

a = -0.2; b = 90; c = -2000;
Δ = b2-4ac
Δ = 902-4·(-0.2)·(-2000)
Δ = 6500
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:

$\sqrt{\Delta}=\sqrt{6500}=\sqrt{100*65}=\sqrt{100}*\sqrt{65}=10\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-10\sqrt{65}}{2*-0.2}=\frac{-90-10\sqrt{65}}{-0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+10\sqrt{65}}{2*-0.2}=\frac{-90+10\sqrt{65}}{-0.4}
$