SAT Question 301faf80: Nonlinear Functions answer and explanation with Desmos solution

Let $x$ represent the first integer and $y$ represent the second integer. If the first integer is $5$ greater than twice the second integer, then $x=2y+5$. It's given that the product of the two integers is $462$; therefore $xy=462$. Substituting $2y+5$ for $x$ in this equation yields $(2y+5)(y)=462$, which can be written as $2y^2+5y=462$. Subtracting $462$ from each side of this equation yields $2y^2+5y-462=0$. The left-hand side of this equation can be factored by finding two values whose product is $2(-462)$, or $-924$, and whose sum is $5$. The two values whose product is $-924$ and whose sum is $5$ are $33$ and $-28$. Thus, the equation $2y^2+5y-462=0$ can be rewritten as $2y^2-28y+33y-462=0$, which is equivalent to $2y(y-14)+33(y-14)=0$, or $(2y+33)(y-14)=0$. By the zero product property, it follows that $2y+33=0$ or $y-14=0$. Subtracting $33$ from both sides of the equation $2y+33=0$ yields $2y=-33$. Dividing both sides of this equation by $2$ yields $y=-\frac{33}{2}$. Since $y$ is a positive integer, the value of $y$ isn't $-\frac{33}{2}$. Adding $14$ to both sides of the equation $y-14=0$ yields $y=14$. Substituting $14$ for $y$ in the equation $xy=462$ yields $x(14)=462$. Dividing both sides of this equation by $14$ yields $x=33$. Therefore, the two integers are $14$ and $33$, so the smaller of the two integers is $14$.