0.2x^2-90x+2000=0

Solution for 0.2x^2-90x+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}
$