P=-0.1x^2+16x-100

Solution for P=-0.1x^2+16x-100 equation:

=-0.1P^2+16P-100

We move all terms to the left:

-(-0.1P^2+16P-100)=0

We get rid of parentheses

0.1P^2-16P+100=0

a = 0.1; b = -16; c = +100;
Δ = b2-4ac
Δ = -162-4·0.1·100
Δ = 216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-6\sqrt{6}}{2*0.1}=\frac{16-6\sqrt{6}}{0.2}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+6\sqrt{6}}{2*0.1}=\frac{16+6\sqrt{6}}{0.2}
$