we have
[tex]\frac{2}{x-2}+ \frac{7}{x^{2}-4}= \frac{5}{x}[/tex]
Remember that
[tex]x^{2}-4=(x+2)(x-2)[/tex]
Multiply by [tex][x(x^{2} -4)][/tex] both sides
[tex][x(x^{2} -4)]\frac{2}{x-2}+[x(x^{2} -4)]\frac{7}{x^{2}-4}= [x(x^{2} -4)]\frac{5}{x}[/tex]
Simplify
[tex]2x(x+2)+7x=5(x^{2}-4) \\ \\2x^{2}+4x+7x=5 x^{2}-20 \\ \\3x^{2}-11x-20=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]3x^{2} -11x-20=0[/tex]
so
[tex]a=3\\b=-11\\c=-20[/tex]
substitute in the formula
[tex]x=\frac{11(+/-)\sqrt{-11^{2}-4(3)(-20)}} {2*3}[/tex]
[tex]x=\frac{11(+/-)\sqrt{121+240}} {6}[/tex]
[tex]x=\frac{11(+/-)\sqrt{361}} {6}[/tex]
[tex]x=\frac{11(+/-)19}{6}[/tex]
[tex]x=\frac{11+19}{6}=5[/tex]
[tex]x=\frac{11-19}{6}=-\frac{4}{3} [/tex]
therefore
the answer is the option
x = negative 4 over 3 and x = 5
