Answer
$x=-1±\sqrt {5}$
Work Step by Step
$\frac{5}{x-3} + \frac{x}{x+3} = \frac{19}{x^2-9}$
$(x-3)(x+3)*\frac{5}{x-3} + (x-3)(x+3)*\frac{x}{x+3} = (x-3)(x+3)*\frac{19}{(x-3)(x+3)}$
$(x+3)*5+(x-3)*x=19$
$5x+15+x^2-3x=19$
$x^2+2x+15=19$
$x^2+2x-4=0$
$x=(-b±\sqrt {b^2-4ac})/2a$
$x=(-2±\sqrt {2^2-4*1*-4})/2*1$
$x=(-2±\sqrt {4+16})/2$
$x=(-2±\sqrt {20})/2$
$x=(-2±\sqrt {4*5})/2$
$x=(-2±2\sqrt {5})/2$
$x=(-1±\sqrt {5})$
Neither answer for $x$ would make the denominators negative, so both answers are valid.