SAT Question 7bd10ef3: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Choice A is correct. The number of solutions to any quadratic equation in the form $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants, can be found by evaluating the expression $b^2-4ac$, which is called the discriminant. If the value of $b^2-4ac$ is a positive number, then there will be exactly two real solutions to the equation. If the value of $b^2-4ac$ is zero, then there will be exactly one real solution to the equation. Finally, if the value of $b^2-4ac$ is negative, then there will be no real solutions to the equation. The given equation $2x^2-4x=t$ is a quadratic equation in one variable, where $t$ is a constant. Subtracting $t$ from both sides of the equation gives $2x^2-4x-t=0$. In this form, $a=2$, $b=-4$, and $c=-t$. The values of $t$ for which the equation has no real solutions are the values of $t$ for which the discriminant $(-4{)}^2-4(2)(-t)$ is negative. $(-4{)}^2-4(2)(-t)<0$ gives $16+8t<0$, or $t<-2$. Of the given choices, only $-3$ is less than $-2$.