x^2+2x-2048=0

Solution for x^2+2x-2048=0 equation:

x^2+2x-2048=0

a = 1; b = 2; c = -2048;
Δ = b2-4ac
Δ = 22-4·1·(-2048)
Δ = 8196
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:

$\sqrt{\Delta}=\sqrt{8196}=\sqrt{4*2049}=\sqrt{4}*\sqrt{2049}=2\sqrt{2049}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{2049}}{2*1}=\frac{-2-2\sqrt{2049}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{2049}}{2*1}=\frac{-2+2\sqrt{2049}}{2}
$