4042+(4042-x)x=2

Solution for 4042+(4042-x)x=2 equation:

4042+(4042-x)x=2

We move all terms to the left:

4042+(4042-x)x-(2)=0

We add all the numbers together, and all the variables

(-1x+4042)x+4042-2=0

We add all the numbers together, and all the variables

(-1x+4042)x+4040=0

We multiply parentheses

-1x^2+4042x+4040=0

a = -1; b = 4042; c = +4040;
Δ = b2-4ac
Δ = 40422-4·(-1)·4040
Δ = 16353924
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:

$\sqrt{\Delta}=\sqrt{16353924}=\sqrt{4*4088481}=\sqrt{4}*\sqrt{4088481}=2\sqrt{4088481}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4042)-2\sqrt{4088481}}{2*-1}=\frac{-4042-2\sqrt{4088481}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4042)+2\sqrt{4088481}}{2*-1}=\frac{-4042+2\sqrt{4088481}}{-2}
$