0=-0.2x^2+41x-1600

Solution for 0=-0.2x^2+41x-1600 equation:

0=-0.2x^2+41x-1600

We move all terms to the left:

0-(-0.2x^2+41x-1600)=0

We add all the numbers together, and all the variables

-(-0.2x^2+41x-1600)=0

We get rid of parentheses

0.2x^2-41x+1600=0

a = 0.2; b = -41; c = +1600;
Δ = b2-4ac
Δ = -412-4·0.2·1600
Δ = 401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-\sqrt{401}}{2*0.2}=\frac{41-\sqrt{401}}{0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+\sqrt{401}}{2*0.2}=\frac{41+\sqrt{401}}{0.4}
$