SAT Question 568aaf27: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Choice D is correct. Substituting $x^2$ from the second equation for $y$ in the first equation yields $x+x^2=12$. Subtracting $12$ from both sides of this equation and rewriting the equation results in $x^2+x-12=0$. Factoring the left-hand side of this equation yields $(x-3)\times (x+4)=0$. Using the zero product property to solve for $x$, it follows that $x-3=0$ and $x+4=0$. Solving each equation for $x$ yields $x=3$ and $x=-4$, respectively. Thus, two possible values of $x$ are $3$ and $-4$. Of the choices given, $3$ is the only possible value of $x$.