SAT Question 03ff48d2: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

An equation of the form $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants, has no real solutions if and only if its discriminant, $b^2-4ac$, is negative. Applying the distributive property to the left-hand side of the equation $x\left(kx-56\right)=-16$ yields $k{x}^2-56x=-16$. Adding $16$ to each side of this equation yields $k{x}^2-56x+16=0$. Substituting $k$ for $a$, $-56$ for $b$, and $16$ for $c$ in $b^2-4ac$ yields a discriminant of ${\left(-56\right)}^2-4\left(k\right)\left(16\right)$, or $3,136-64k$. If the given equation has no real solution, it follows that the value of $3,136-64k$ must be negative. Therefore, $3,136-64k<0$. Adding $64k$ to both sides of this inequality yields $3,136<64k$. Dividing both sides of this inequality by $64$ yields $49<k$, or $k>49$. Since it's given that $k$ is an integer, the least possible value of $k$ is $50$.