P(x)=-x^2+1000x-10x-3400

Solution for P(x)=-x^2+1000x-10x-3400 equation:

(P)=-P^2+1000P-10P-3400

We move all terms to the left:

(P)-(-P^2+1000P-10P-3400)=0

We get rid of parentheses

P^2-1000P+10P+P+3400=0

We add all the numbers together, and all the variables

P^2-989P+3400=0

a = 1; b = -989; c = +3400;
Δ = b2-4ac
Δ = -9892-4·1·3400
Δ = 964521
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:

$\sqrt{\Delta}=\sqrt{964521}=\sqrt{9*107169}=\sqrt{9}*\sqrt{107169}=3\sqrt{107169}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-989)-3\sqrt{107169}}{2*1}=\frac{989-3\sqrt{107169}}{2}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-989)+3\sqrt{107169}}{2*1}=\frac{989+3\sqrt{107169}}{2}
$