It is possible that this equation was meant to be:
$$\frac{14}{x^2-9}+\frac{6}{x-3}=\frac{8}{x+3}$$
in which case, multiply all terms by x2 -9
$$14+\frac{6(x^2-9)}{x-3}=\frac{8(x^2-9))}{x+3}$$
Express x2 - 9 as (x+3)(x-3) and cancel as appropriate
$$14+6(x+3)=8(x-3)$$
Expand bracketed terms and collect like terms on the same side
$$14+18+24=8x-6x$$
$$56=2x$$
$$x=28$$
.
$${\frac{{\mathtt{14}}}{{{\mathtt{x}}}^{{\mathtt{2}}}}}{\mathtt{\,-\,}}{\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{6}}}{{\mathtt{x}}}}{\mathtt{\,-\,}}{\mathtt{3}} = {\frac{{\mathtt{8}}}{{\mathtt{x}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{211}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{15}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{211}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{15}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{1.035\: \!055\: \!936\: \!422\: \!263\: \!3}}\\
{\mathtt{x}} = {\mathtt{0.901\: \!722\: \!603\: \!088\: \!93}}\\
\end{array} \right\}$$
happy 7 is correct.
If you need to show some work: 14/x² - 9 + 6/x - 3 = 8/x + 3
Multiply each term by x² to get rid of denominators:
(x²)(14/x²) - (x²)(9) + (x²)(6/x) - (x²)(3) = (x²)(8/x) + (x²)(3)
14 - 9x² + 6x - 3x² = 8x + 3x²
14 - 12x² + 6x = 8x + 3x²
14 - 15x² - 2x = 0
-15x² - 2x + 14 = 0
15x² + 2x - 14 = 0
Using the quadratic equation: x = [ -b ± √(b² -4ac) ] / (2a)
a = 15 b = 2 c = -14
x = [ -2 ± √(2² -4·15·-14) ] / (2·15)
x = [ -2 ± √(4 + 840) ] / (30)
x = [ -2 ± √(844) ] / (30)
x = [ -2 ± √4√211 ] / 30
x = [ -2 ± 2√211 ] / 30
x = [ -1 ± √211 ] / 15
It is possible that this equation was meant to be:
$$\frac{14}{x^2-9}+\frac{6}{x-3}=\frac{8}{x+3}$$
in which case, multiply all terms by x2 -9
$$14+\frac{6(x^2-9)}{x-3}=\frac{8(x^2-9))}{x+3}$$
Express x2 - 9 as (x+3)(x-3) and cancel as appropriate
$$14+6(x+3)=8(x-3)$$
Expand bracketed terms and collect like terms on the same side
$$14+18+24=8x-6x$$
$$56=2x$$
$$x=28$$
.