2x+x^2=440

Solution for 2x+x^2=440 equation:

2x+x^2=440

We move all terms to the left:

2x+x^2-(440)=0

a = 1; b = 2; c = -440;
Δ = b2-4ac
Δ = 22-4·1·(-440)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-42}{2*1}=\frac{-44}{2}
=-22
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+42}{2*1}=\frac{40}{2}
=20
$