40=10x-0.1x^2

Solution for 40=10x-0.1x^2 equation:

40=10x-0.1x^2

We move all terms to the left:

40-(10x-0.1x^2)=0

We get rid of parentheses

0.1x^2-10x+40=0

a = 0.1; b = -10; c = +40;
Δ = b2-4ac
Δ = -102-4·0.1·40
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{21}}{2*0.1}=\frac{10-2\sqrt{21}}{0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{21}}{2*0.1}=\frac{10+2\sqrt{21}}{0.2}
$