P(x)=0.004x^2+2.8x-50

Solution for P(x)=0.004x^2+2.8x-50 equation:

(P)=0.004P^2+2.8P-50

We move all terms to the left:

(P)-(0.004P^2+2.8P-50)=0

We get rid of parentheses

-0.004P^2+P-2.8P+50=0

We add all the numbers together, and all the variables

-0.004P^2-1.8P+50=0

a = -0.004; b = -1.8; c = +50;
Δ = b2-4ac
Δ = -1.82-4·(-0.004)·50
Δ = 4.04
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{-(-1.8)-\sqrt{4.04}}{2*-0.004}=\frac{1.8-\sqrt{4.04}}{-0.008}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.8)+\sqrt{4.04}}{2*-0.004}=\frac{1.8+\sqrt{4.04}}{-0.008}
$