Решите уравнение)))) 10/x+4/(x-3)=1

Решите задачу:

$\frac{10}{x}+\frac{4}{x-3}=1;\ x\neq0\ \wedge\ x-3\neq0\to x\neq0\ \wedge\ x\neq3\\\\\frac{10(x-3)}{x(x-3)}+\frac{4x}{x(x-3)}=1\\\\\frac{10x-30+4x}{x^2-3x}=1\\\\\frac{14x-30}{x^2-3x}=1\iff14x-30=x^2-3x\\\\x^2-3x-14x+30=0\\\\x^2-17x+30=0$

$x^2-2x-15x+30=0\\\\x(x-2)-15(x-2)=0\\\\(x-2)(x-15)=0\iff x-2=0\ \vee\ x-15=0\\\\x=2\ \vee\ x=15$