SAT Question 6011a3f8: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Choice A is correct. A quadratic equation of the form $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants, has either no real solutions, exactly one real solution, or exactly two real solutions. That is, for the given equation to have more than one real solution, it must have exactly two real solutions. When the value of the discriminant, or $b^2-4ac$, is greater than 0, the given equation has exactly two real solutions. In the given equation, $64x^2+bx+25=0$, $a=64$ and $c=25$. Therefore, the given equation has exactly two real solutions when ${\left(b\right)}^2-4\left(64\right)\left(25\right)>0$, or $b^2-6,400>0$. Adding $6,400$ to both sides of this inequality yields $b^2>6,400$. Taking the square root of both sides of $b^2>6,400$ yields two possible inequalities: $b<-80$ or $b>80$. Of the choices, only choice A satisfies $b<-80$ or $b>80$.