200=0.1x^2+16x

Solution for 200=0.1x^2+16x equation:

200=0.1x^2+16x

We move all terms to the left:

200-(0.1x^2+16x)=0

We get rid of parentheses

-0.1x^2-16x+200=0

a = -0.1; b = -16; c = +200;
Δ = b2-4ac
Δ = -162-4·(-0.1)·200
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{21}}{2*-0.1}=\frac{16-4\sqrt{21}}{-0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{21}}{2*-0.1}=\frac{16+4\sqrt{21}}{-0.2}
$