Quadratic distinct real solutions count practice

A square of a real number cannot equal $-16$, so there are $0$ distinct real solutions.

The equation is equivalent to $x^2 = -12$. A square of a real number cannot be negative, so there are $0$ distinct real solutions.

The trinomial is a perfect square: $x^2 - 10x + 25 = (x - 5)^2 = 0$. There is exactly $1$ distinct real solution, $x = 5$.

The equation is already written as a square equal to zero, so there is exactly $1$ distinct real solution, $x = -7$.

By the zero-product property, $x - 8 = 0$ or $x + 3 = 0$, so $x = 8$ or $x = -3$. There are $2$ distinct real solutions.

Subtracting $15$ gives $3(x + 2)^2 = -15$, so $(x + 2)^2 = -5$. A square of a real number cannot be negative, so there are $0$ distinct real solutions.

The trinomial is a perfect square: $9x^2 - 30x + 25 = (3x - 5)^2 = 0$. There is exactly $1$ distinct real solution, $x = \dfrac{5}{3}$.

The quadratic factors as $(2x + 1)(x - 3) = 0$, giving $x = -\dfrac{1}{2}$ and $x = 3$. There are $2$ distinct real solutions.