280=16x-0.1x^2

Solution for 280=16x-0.1x^2 equation:

280=16x-0.1x^2

We move all terms to the left:

280-(16x-0.1x^2)=0

We get rid of parentheses

0.1x^2-16x+280=0

a = 0.1; b = -16; c = +280;
Δ = b2-4ac
Δ = -162-4·0.1·280
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-12}{2*0.1}=\frac{4}{0.2}
=20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+12}{2*0.1}=\frac{28}{0.2}
=140
$