Solving quadratics (leading coefficient not 1). Game on.

The quadratic factors as $(2x + 5)(x - 3) = 0\textsf{.}$ By the zero-product property, $2x + 5 = 0$ or $x - 3 = 0\textsf{,}$ so $x = -\dfrac{5}{2}$ or $x = 3\textsf{.}$ The positive solution is $3\textsf{.}$

The quadratic factors as $(4x + 1)(x - 6) = 0\textsf{.}$ By the zero-product property, $4x + 1 = 0$ or $x - 6 = 0\textsf{,}$ so $x = -\dfrac{1}{4}$ or $x = 6\textsf{.}$ The positive solution is $6\textsf{.}$

The quadratic factors as $(6x + 5)(x - 2) = 0\textsf{.}$ By the zero-product property, $6x + 5 = 0$ or $x - 2 = 0\textsf{,}$ so $x = -\dfrac{5}{6}$ or $x = 2\textsf{.}$ The positive solution is $2\textsf{.}$

The quadratic factors as $(3x + 2)(x - 7) = 0\textsf{.}$ By the zero-product property, $3x + 2 = 0$ or $x - 7 = 0\textsf{,}$ so $x = -\dfrac{2}{3}$ or $x = 7\textsf{.}$ The positive solution is $7\textsf{.}$

The quadratic factors as $(5x + 2)(x - 4) = 0\textsf{.}$ By the zero-product property, $5x + 2 = 0$ or $x - 4 = 0\textsf{,}$ so $x = -\dfrac{2}{5}$ or $x = 4\textsf{.}$ The positive solution is $4\textsf{.}$

The quadratic factors as $(2x + 7)(x - 5) = 0\textsf{.}$ By the zero-product property, $2x + 7 = 0$ or $x - 5 = 0\textsf{,}$ so $x = -\dfrac{7}{2}$ or $x = 5\textsf{.}$ The positive solution is $5\textsf{.}$

The quadratic factors as $(4x + 3)(x - 3) = 0\textsf{.}$ By the zero-product property, $4x + 3 = 0$ or $x - 3 = 0\textsf{,}$ so $x = -\dfrac{3}{4}$ or $x = 3\textsf{.}$ The positive solution is $3\textsf{.}$

The quadratic factors as $(5x + 4)(x - 3) = 0\textsf{.}$ By the zero-product property, $5x + 4 = 0$ or $x - 3 = 0\textsf{,}$ so $x = -\dfrac{4}{5}$ or $x = 3\textsf{.}$ The positive solution is $3\textsf{.}$