SAT Question 46308566: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

A quadratic equation of the form $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants, has no real solution if and only if its discriminant, $b^2-4ac$, is negative. It’s given that in the equation $r^2+qr=2r-55$, $q$ is an integer constant. Subtracting $2r$ and adding $55$ to both sides of this equation yields $r^2+qr-2r+55=0$, or $r^2+(q-2)r+55=0$. In this equation, $a=1$, $b=q-2$, and $c=55$. Substituting $1$ for $a$, $q-2$ for $b$, and $55$ for $c$ in the discriminant expression yields $(q-2{)}^2-4(1)(55)$, or $(q-2{)}^2-220$. Since the value of the discriminant must be negative, it follows that $(q-2{)}^2-220<0$, or $(q-2{)}^2<220$. This is equivalent to $\left|q-2\right|<\sqrt{220}$, or $-\sqrt{220}<q-2<\sqrt{220}$. Adding $2$ to each part of this inequality yields $2-\sqrt{220}<q<2+\sqrt{220}$. Since $\sqrt{220}$ is approximately $14.83$, it follows that $-12.83<q<16.83$. Since $q$ is an integer, the largest possible value of $q$ is $16$.