sat-dog

The correct answer is $-1$. A quadratic equation in the form $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants, has exactly one solution when its discriminant, $b^2-4ac$, is equal to $0$. In the given equation, $-16x^2-8x+c=0$, $a=-16$ and $b=-8$. Substituting $-16$ for $a$ and $-8$ for $b$ in $b^2-4ac$ yields ${\left(-8\right)}^2-4\left(-16\right)\left(c\right)$, or $64+64c$. Since the given equation has exactly one solution, $64+64c=0$. Subtracting $64$ from both sides of this equation yields $64c=-64$. Dividing both sides of this equation by $64$ yields $c=-1$. Therefore, the value of $c$ is $-1$.