Quadratic distinct real solutions count. Game on.

The equation is already $(x - 4)^{2} = -16\textsf{.}$ A real number squared cannot be negative, so there are $0$ distinct real solutions.

Subtract $12$: $x^{2} = -12\textsf{.}$ A real number squared cannot be negative, so there are $0$ distinct real solutions.

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

The equation is already a square equal to zero: $(x + 7)^{2} = 0\textsf{,}$ so $x = -7\textsf{.}$ There is exactly $1$ distinct real solution.

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

Subtract $15$: $3(x + 2)^{2} = -15\textsf{.}$ Divide by $3$: $(x + 2)^{2} = -5\textsf{.}$ A real number squared cannot be negative, so there are $0$ distinct real solutions.

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

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