P(x)=0.02x^2+38x-5250

Solution for P(x)=0.02x^2+38x-5250 equation:

(P)=0.02P^2+38P-5250

We move all terms to the left:

(P)-(0.02P^2+38P-5250)=0

We get rid of parentheses

-0.02P^2+P-38P+5250=0

We add all the numbers together, and all the variables

-0.02P^2-37P+5250=0

a = -0.02; b = -37; c = +5250;
Δ = b2-4ac
Δ = -372-4·(-0.02)·5250
Δ = 1789
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{-(-37)-\sqrt{1789}}{2*-0.02}=\frac{37-\sqrt{1789}}{-0.04}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-37)+\sqrt{1789}}{2*-0.02}=\frac{37+\sqrt{1789}}{-0.04}
$