x^2+2x=2304

Solution for x^2+2x=2304 equation:

x^2+2x=2304

We move all terms to the left:

x^2+2x-(2304)=0

a = 1; b = 2; c = -2304;
Δ = b2-4ac
Δ = 22-4·1·(-2304)
Δ = 9220
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:

$\sqrt{\Delta}=\sqrt{9220}=\sqrt{4*2305}=\sqrt{4}*\sqrt{2305}=2\sqrt{2305}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{2305}}{2*1}=\frac{-2-2\sqrt{2305}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{2305}}{2*1}=\frac{-2+2\sqrt{2305}}{2}
$