Algebra Question that I can't figure out

solve it using the quadratic formula.

x^2 + x + 1 = 0

a = 1, b = 1, c = 1.

x = [ -b +/- sqrt(b^2 - 4ac) ] / [ 2a ]
x = [ -1 +/- sqrt(1^2 - 4(1)(1)) ] / [ 2*1]
x = [ -1 +/- sqrt(1 - 4) ] / 2
x = [ -1 +/- sqrt(-3) ] / 2

So you have two imaginary roots.

It shouldn't matter whether you use the quadratic formula or complete the square. Here, let me show you what happens when completing the square.

x^2 + x + 1 = 0

Add (1/4) to both sides,

x^2 + x + 1/4 + 1 = 1/4
(x + [1/2])^2 + 1 = 1/4
(x + [1/2])^2 = (1/4) - 1
(x + [1/2])^2 = (-3/4)
x + [1/2] = +/- sqrt(-3/4)
x + [1/2] = +/- sqrt(-3)/2
x = [-1/2] +/- sqrt(-3)/2
x = (-1 +/- sqrt(-3))/2

LostGirl:

x^2-x+1=0

Dee has answered the question well. I just want to add a little.

The bit under the square root in the quadratic formula is called the discriminate. (The symbol for it is a triangle)
The discriminant = b ² - 4ac
The reason it is called the discriminant is that it discriminates between the types of roots that the equation will have.
If the discriminant is > 0 the equation will have 2 real roots.
If the discriminant is = 0 the equation will have 1 real roots.
If the discriminant is < 0 the equation will have 0 real roots.
If the discriminant is a perfect square then the equation will have rational roots.

for your equation, a=1, b=-1 and c=1
discriminant = 1 - 4 = -3 < 0
therefore the equation will have no real solution.

If you were to graph y = x ²-x+1
then, when y =0, there would be no solution which means that the graph does not cross the x axis.
This is the graph of a concave up parabola that sits totally above the x axis.
I hope you can understand all this.