SAT Question 6aefc52b: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

It's given by the first equation of the system of equations that $y=-2.5$. Substituting $-2.5$ for $y$ in the second given equation, $y=x^2+8x+k$, yields $-2.5=x^2+8x+k$. Adding $2.5$ to both sides of this equation yields $0=x^2+8x+k+2.5$. A quadratic equation of the form $0=a{x}^2+bx+c$, where $a$, $b$, and $c$ are constants, has no real solutions if and only if its discriminant, $b^2-4ac$, is negative. In the equation $0=x^2+8x+k+2.5$, where $k$ is a positive integer constant, $a=1$, $b=8$, and $c=k+2.5$. Substituting $1$ for $a$, $8$ for $b$, and $k+2.5$ for $c$ in $b^2-4ac$ yields $8^2-4\left(1\right)\left(k+2.5\right)$, or $64-4\left(k+2.5\right)$. Since this value must be negative, $64-4\left(k+2.5\right)<0$. Adding $4\left(k+2.5\right)$ to both sides of this inequality yields $64<4\left(k+2.5\right)$. Dividing both sides of this inequality by $4$ yields $16<k+2.5$. Subtracting $2.5$ from both sides of this inequality yields $13.5<k$. Since $k$ is a positive integer constant, the least possible value of $k$ is $14$.