A problem by charishma t

let ( $x^{2}$ +x+1)=t then t(t+1)=12 by solving this t=-4, t=3 ( $x^{2}$ +x+1)=-4 $x^{2}$ +x+5=0 imaginary roots $b^{2}$ -4ac <0 $x^{2}$ +x+1=3 by solving this x=-2,x=1 two real roots

Denote $x^2 + x = a$ , then substituting into the original equation,

$(a + 1)(a + 2) = a^2 + 3a + 2 = 12 \:\Rightarrow\: a \in \{-5, 2\}$

We now have two quadratic equations, $x^2 + x - 2 = 0$ , and $x^2 + x + 5 = 0$ . The discriminants of these polynomials are, respectively, $10$ and $-19$ . Thus, there are 2 unique real solutions for the former equation ( $1$ and $-1$ ), and no real solutions for the latter.

The answer is then $\boxed{2}$

x is either 1 or -2. :D

Take $x^2 + x = t$ . We have $(t+1)(t+2) = 12 \Rightarrow t^2 + 3y - 10 = 0$ . This quadratic equation in $t$ has two solutions: 2 or -5. As according to what we originally defined $t$ to be, we have $x^2 + x = 2$ or $x^2 + x = -5$ . The former has 2 real roots and the latter has 0 real roots using the discriminant. And so we conclude the number of roots to be $2+0=\boxed{2}$ .