Answer
$ x=-\frac{3}{2}$ or $x=5$
Work Step by Step
$$\frac{10}{x}-\frac{12}{x-3}+4=0$$,
First multiply both sides by $x(x-3)$:
$x(x-3)\frac{10}{x}-x(x-3)\frac{12}{x-3}+x(x-3)4=x(x-3)0$,
$10(x-3)-12x+4x(x-3)=0$,
And then distribute:
$10(x-3)-12x+4x(x-3)=0$,
$10x-30-12x+4x^2-12x=0$,
And then combine like terms:
$10x-30-12x+4x^2-12x=0$,
$4x^2+10x-12x-12x-30=0$,
$4x^2-14x-30=0$,
And then factor out and solve for trinomial:
$4x^2-20x+6x-30=0$,
$4x(x-5)+6(x-5)=0$,
$(4x+6)(x-5)=0$,
either $4x+6=0, x=-\frac{3}{2}$ or $x-5=0, x=5$
