The discriminant and quadratic formula. Game on.

Here $a = 1$, $b = -5$, and $c = 6$. Substituting into the quadratic formula gives $x = \dfrac{-(-5) \pm \sqrt{(-5)^{2} - 4(1)(6)}}{2(1)} = \dfrac{5 \pm 1}{2}$, so $x = 2$ and $x = 3$.

Here $a = 2$, $b = 3$, and $c = -5$. Substituting into $b^{2} - 4ac$ gives $3^{2} - 4(2)(-5) = 9 + 40 = 49$.

Here $a = 1$, $b = 4$, and $c = 1$. Substituting into the quadratic formula gives $x = \dfrac{-4 \pm \sqrt{4^{2} - 4(1)(1)}}{2(1)} = \dfrac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}$.

Here $a = 1$, $b = 6$, and $c = 9$. Substituting into $b^{2} - 4ac$ gives $6^{2} - 4(1)(9) = 0$, so there is exactly $1$ distinct real solution.

Here $a = 1$, $b = 1$, and $c = 1$. Substituting into $b^{2} - 4ac$ gives $1^{2} - 4(1)(1) = -3$, so there are $0$ real solutions.

Here $a = 2$, $b = -7$, and $c = 3$. Substituting into $b^{2} - 4ac$ gives $(-7)^{2} - 4(2)(3) = 49 - 24 = 25$.

Here $a = 1$, $b = 0$, and $c = -16$. Substituting into $b^{2} - 4ac$ gives $0^{2} - 4(1)(-16) = 64$.

Here $a = 1$, $b = -4$, and $c = 4$. Substituting into $b^{2} - 4ac$ gives $(-4)^{2} - 4(1)(4) = 0$, so there is exactly $1$ distinct real solution.