200x+x^2-1544=0

Solution for 200x+x^2-1544=0 equation:

200x+x^2-1544=0

a = 1; b = 200; c = -1544;
Δ = b2-4ac
Δ = 2002-4·1·(-1544)
Δ = 46176
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:

$\sqrt{\Delta}=\sqrt{46176}=\sqrt{16*2886}=\sqrt{16}*\sqrt{2886}=4\sqrt{2886}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-4\sqrt{2886}}{2*1}=\frac{-200-4\sqrt{2886}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+4\sqrt{2886}}{2*1}=\frac{-200+4\sqrt{2886}}{2}
$