SAT Question 1fe32f7d: 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, $-4ac+b^2$, is negative. In the given equation, $a=-1$ and $c=-676$. Substituting $-1$ for $a$ and $-676$ for $c$ in this expression yields a discriminant of $b^2-4\left(-1\right)\left(-676\right)$, or $b^2-2,704$. Since this value must be negative, $b^2-2,704<0$, or $b^2<2,704$. Taking the positive square root of each side of this inequality yields $b<52$. Since $b$ is a positive integer, and the greatest integer less than $52$ is $51$, the greatest possible value of $b$ is $51$.