1300-2x=400x+x^2

Solution for 1300-2x=400x+x^2 equation:

1300-2x=400x+x^2

We move all terms to the left:

1300-2x-(400x+x^2)=0

We get rid of parentheses

-x^2-400x-2x+1300=0

We add all the numbers together, and all the variables

-1x^2-402x+1300=0

a = -1; b = -402; c = +1300;
Δ = b2-4ac
Δ = -4022-4·(-1)·1300
Δ = 166804
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{166804}=\sqrt{4*41701}=\sqrt{4}*\sqrt{41701}=2\sqrt{41701}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-402)-2\sqrt{41701}}{2*-1}=\frac{402-2\sqrt{41701}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-402)+2\sqrt{41701}}{2*-1}=\frac{402+2\sqrt{41701}}{-2}
$