[tex]\dfrac{2x-3}x = \dfrac{14}{x+5}\\\\\implies (2x-3)(x+5) = 14x\\\\\implies 2x^2 +10x -3x -15 -14x=0\\\\\implies 2x^2-7x-15=0\\\\\text{Apply quadratic formula,}~~x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}.\\\\\text{In this case,}~ a=2, b=-7~ \text{and}~ c = -15. \\\\\\x = \dfrac{-(-7) \pm \sqrt{(-7)^2 -4\cdot 2 \cdot (-15)}}{2(2)}\\\\\\\implies x = \dfrac{7 \pm \sqrt{169}}{4}\\\\\\\implies x = \dfrac{7 \pm 13}4\\\\\\[/tex]
[tex]\text{Hence} ~x = \dfrac{7+13}{4} = \dfrac{20}4 = 5 ~\text{or}~ x = \dfrac{7-13}4 = -\dfrac{6}4 = -\dfrac 32[/tex]
