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}#

#b^2-4ac#, 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}#