# How do you solve x^2+x+1=0?

Apr 11, 2015

You can use the standard formula which allows you to solve any quadratic equation, which is

${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

${b}^{2} - 4 a c$, also called $\setminus \Delta$ (delta), is the discriminant: if $\setminus \Delta \setminus \ge 0$, then the solutions are real numbers (you have two distinct solutions if $\setminus \Delta > 0$ and two solutions collapsed into the same point if $\setminus \Delta = 0$), while if $\setminus \Delta < 0$, the two solutions are complex and conjugated.

In your case, the general equation $a {x}^{2} + b x + c$ translates into ${x}^{2} + x + 1$ if $a = b = c = 1$. Plugging these values into the solving formula written at the beginning, you have

${x}_{1 , 2} = \setminus \frac{- 1 \setminus \pm \setminus \sqrt{{1}^{2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = - \frac{1}{2} \setminus \pm \setminus \frac{\sqrt{- 3}}{2}$

Since the discriminant is $- 3$, there are no real solutions. If you are allowed to use complex numbers, you'll see that $\setminus \sqrt{- 3} = i \setminus \sqrt{3}$, and so the two conjugated solutions are ${x}_{1} = - \frac{1}{2} + i \setminus \frac{\setminus \sqrt{3}}{2}$, and ${x}_{2} = - \frac{1}{2} - i \setminus \frac{\setminus \sqrt{3}}{2}$