Solve for x \(\frac{x^2+x+1}{x+1}=x+2\)
\(\frac{x^2 +x+1}{x+1}=x+2 \\ x^2 + x + 1 = (x+2)(x+1) \\ x^2 +x+1=x^2+3x+2 \\ 0 = x^2 +3x +2 -(x^2 +x+1) \\ 0 = x^2 +3x +2 -x^2 -x-1 \\ 0 = 2x+1 \\ -1 = 2x \\ -\frac{1}{2}=x\)
