Solution:-
[tex] - (x + 3) + \frac{3}{4x} + 5 = 0[/tex]
[tex]⇒ - x - 3 + \frac{3}{4x} + 5 = 0[/tex]
[tex]⇒ - x + \frac{3}{4x} = - 5 + 3[/tex]
[tex]⇒ \frac{ - 4 {x}^{2} + 3 }{4x} = - 2[/tex]
[tex]⇒ - 4 {x}^{2} + 3 = - 8x[/tex]
[tex]⇒ - 4 {x}^{2} + 8x + 3 = 0[/tex]
[tex]⇒4 {x}^{2} - 8x - 3 = 0[/tex]
[tex]⇒x = \frac{ - ( - 8) \pm\sqrt{( - 8)^{2} - 4 \times 4 \times ( - 3) } }{2 \times 4} [/tex]
[tex]⇒x = \frac{8 \pm \sqrt{64 + 48} }{8} [/tex]
[tex]⇒x = \frac{8 \pm \sqrt{112} }{8} [/tex]
[tex]⇒ x_{1} = \frac{8 + \sqrt{112} }{8} \: \: ,\: \: x_{2} = \frac{8 - \sqrt{112} }{8} [/tex]
[tex]⇒ x_{1} = \frac{8 + 10.58}{8} \: \: , \: \: x_{2} = \frac{8 - 10.58}{8} [/tex]
[tex]⇒ x_{1} = \frac{18.58}{8} \: \: , \: \: x_{2} = \frac{ - 2.58}{8} [/tex]
[tex]⇒ x_{1} = 2.32\: \: , \: \: x_{2} = - 0.32[/tex]
[tex] \sf \red{ \therefore{x_{1} = 2.32\: \: and\: \: x_{2} = - 0.32}}[/tex]
