Note, original question details are here:

Oh, for sure, it can have real roots. You can express it as [math]ax^2 = -c -bx[/math]

So, you are looking for an intersection of a quadratic with a straight line. It’s often going to have solutions.

By the Quadratic formula, the requirement for it to have real roots is [math]b^2 >= 4ac[/math]. So for instance, if a and c are 1, then [math]x^2 + bx + 1 = 0[/math] has real solutions for [math]b >= 2[/math]

The solution for [math]x^2 + 2x + 1 = 0 [/math] is x = -1. But if you put b less than 2, it’s got no solution. This graph using the Desmos Graphing Calculator may help you see how it works, for that case, re-arranged as [math]x^2 = -2x - 1[/math]

Notice that the graphs meet at x = -1.

And here it is for the equation [math]x^2 = -x - 1[/math], a rearrangement of the equation [math]x^2 + x + 1 = 0 [/math]which has no real solution.