x^2+2x-1640=0

Solution for x^2+2x-1640=0 equation:

x^2+2x-1640=0

a = 1; b = 2; c = -1640;
Δ = b2-4ac
Δ = 22-4·1·(-1640)
Δ = 6564
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:

$\sqrt{\Delta}=\sqrt{6564}=\sqrt{4*1641}=\sqrt{4}*\sqrt{1641}=2\sqrt{1641}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{1641}}{2*1}=\frac{-2-2\sqrt{1641}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{1641}}{2*1}=\frac{-2+2\sqrt{1641}}{2}
$