P=(-0.001x^2)+10x

Solution for P=(-0.001x^2)+10x equation:

=(-0.001P^2)+10P

We move all terms to the left:

-((-0.001P^2)+10P)=0


We calculate terms in parentheses: -((-0.001P^2)+10P), so:

(-0.001P^2)+10P

We get rid of parentheses

-0.001P^2+10P

Back to the equation:

-(-0.001P^2+10P)


We get rid of parentheses

0.001P^2-10P=0

a = 0.001; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·0.001·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*0.001}=\frac{0}{0.002}
=0
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*0.001}=\frac{20}{0.002}
=10000
$