Answer:
The values of x are -3 and 5
Step-by-step explanation:
we have
[tex]x-\frac{15}{x}=2[/tex]
Multiply by x both sides to remove the fraction
[tex]x^{2} -15=2x\\\\x^{2} -2x-15=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]x^{2} -2x-15=0[/tex]
so
[tex]a=1\\b=-2\\c=-15[/tex]
substitute in the formula
[tex]x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(1)(-15)}} {2(1)}[/tex]
[tex]x=\frac{2(+/-)\sqrt{64}} {2}[/tex]
[tex]x=\frac{2(+/-)8} {2}[/tex]
[tex]x_1=\frac{2(+)8} {2}=5[/tex]
[tex]x_2=\frac{2(-)8} {2}=-3[/tex]
therefore
The values of x are -3 and 5
