SAT Question 74473be4: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Choice D is correct. The number of solutions to a quadratic equation in the form $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants, can be determined by the value of the discriminant, $b^2-4ac$. If the value of the discriminant is greater than zero, then the quadratic equation has two distinct real solutions. If the value of the discriminant is equal to zero, then the quadratic equation has exactly one real solution. If the value of the discriminant is less than zero, then the quadratic equation has no real solutions. For the quadratic equation in choice D, $5x^2-14x+49=0$, $a=5$, $b=-14$, and $c=49$. Substituting $5$ for $a$, $-14$ for $b$, and $49$ for $c$ in $b^2-4ac$ yields ${\left(-14\right)}^2-4\left(5\right)\left(49\right)$, or $-784$. Since $-784$ is less than zero, it follows that the quadratic equation $5x^2-14x+49=0$ has no real solutions.