P=-0.02n^2+3.40n-16

Solution for P=-0.02n^2+3.40n-16 equation:

=-0.02P^2+3.40P-16

We move all terms to the left:

-(-0.02P^2+3.40P-16)=0

We get rid of parentheses

0.02P^2-3.40P+16=0

We add all the numbers together, and all the variables

0.02P^2-3.4P+16=0

a = 0.02; b = -3.4; c = +16;
Δ = b2-4ac
Δ = -3.42-4·0.02·16
Δ = 10.28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3.4)-\sqrt{10.28}}{2*0.02}=\frac{3.4-\sqrt{10.28}}{0.04}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3.4)+\sqrt{10.28}}{2*0.02}=\frac{3.4+\sqrt{10.28}}{0.04}
$