Question

Solve for value of $x$:
$x^2+x+1<0$

Answer 1

$x^2+x+1<0$

$x^2+x+\frac{1}{2^2}-\frac{1}{2^2}+1<0$

$(x^2+x+\frac{1}{2^2})-\frac{1}{4}+1<0$

${(x+\frac{1}{2})}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2<0$

$(x+\frac{1}{2}-\frac{\sqrt{3}}{2})(x+\frac{1}{2}+\frac{\sqrt{3}}{2})<0$

$(x+\frac{(1-\sqrt{3})}{2})<0$ or $(x+\frac{1+\sqrt{3}}{2})<0$

$x<-\left(\frac{1-\sqrt{3}}{2}\right)$ or $x<-\frac{(1+\sqrt{3})}{2}$

$x<\frac{(\sqrt{3}-1)}{2}$ or $x<-\frac{(1+\sqrt{3})}{2}$

$\therefore x<-\frac{(1+\sqrt{3})}{2}$ $(\frac{\sqrt{3}-1}{2}>-\frac{(1+\sqrt{3})}{2})$.