5 x 2 + 10 x + 16 = 0 How many distinct real solutions does

Choice D is correct. The number of solutions of a quadratic equation of 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 positive, then the quadratic equation has exactly 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 negative, then the quadratic equation has zero real solutions. In the given equation, $5x^2+10x+16=0$, $a= 5$, $b=10$, and $c=16$. Substituting these values for $a$, $b$, and $c$ in $b^2-4ac$ yields ${\left(10\right)}^2-4\left(5\right)\left(16\right)$, or $-220$. Since the value of its discriminant is negative, the given equation has zero real solutions. Therefore, the number of distinct real solutions the given equation has is zero.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.