You can use the standard formula which allows you to solve any quadratic equation, which is
x_{1,2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}x1,2=−b±√b2−4ac2a
b^2-4acb2−4ac, also called \Delta (delta), is the discriminant: if \Delta\ge 0, then the solutions are real numbers (you have two distinct solutions if \Delta > 0 and two solutions collapsed into the same point if \Delta=0), while if \Delta<0, the two solutions are complex and conjugated.
In your case, the general equation ax^2+bx+c translates into x^2+x+1 if a=b=c=1. Plugging these values into the solving formula written at the beginning, you have
x_{1,2} = \frac{-1 \pm \sqrt{1^2-4*1*1}}{2*1} = -1/2 \pm \sqrt{-3}/2
Since the discriminant is -3, there are no real solutions. If you are allowed to use complex numbers, you'll see that \sqrt{-3}=i\sqrt(3), and so the two conjugated solutions are x_1=-1/2 + i\frac{\sqrt{3}}{2}, and x_2=-1/2 - i\frac{\sqrt{3}}{2}
