Q.Solve $x^2+x+1=0$.

Every quadratic equation $ax^2+bx+c=0$ (with $a\neq0$) has solutions given by the familiar formula $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$, and this formula continues to work even when the discriminant $D=b^2-4ac$ is negative or when the coefficients $a,b,c$ are themselves complex — the Fundamental Theorem of Algebra guarantees that a degree-$n$ polynomial equation always has exactly $n$ roots in the complex numbers, so a quadratic is never left without a solution. When $a,b,c$ are real and $D<0$, the square root $\sqrt{D}$ is rewritten as $i\sqrt{|D|}$, and the two roots that come out are complex conjugates of each other — if $p+iq$ is one root, $p-iq$ is automatically the other. When the coefficients are themselves complex (or when the discriminant is a general complex number rather than a negative real one), finding $\sqrt{D}$ requires the square-root-of-a-complex-number technique (writing the square root as $a+ib$ and solving two simultaneous equations), and the two roots need not be conjugates of each other. A related, very useful trick is evaluating a polynomial $p(x)$ at a known complex root: if $x=p+iq$ satisfies a quadratic $x^2+Bx+C=0$ with real coefficients built from that root and its conjugate, then dividing the target polynomial by that quadratic reduces the whole evaluation to a short remainder calculation instead of repeatedly expanding high powers of a complex number.