Помогите решить ,пожалуйста пример №7.34(а)

$ODZ \\ x-3 \neq 0 \\ x \neq 3 \\ \\ \frac{x^2+3x}{2(x-3)}+ \frac{x+12}{6}= \frac{3x}{x-3} \\ \frac{3(x^2+3x)}{3*2(x-3)}+ \frac{(x-3)(x+12)}{6(x-3)}= \frac{6*3x}{6(x-3)} \\ \frac{3x^2+9x}{6(x-3)}+ \frac{x^2+12x-3x-36}{6(x-3)}= \frac{18x}{6(x-3)} \\ \frac{3x^2+9x}{6(x-3)}+ \frac{x^2+9x-36}{6(x-3)}= \frac{18x}{6(x-3)} \\ 3x^2+9x+x^2+9x-36=18x \\ 4x^2+18x-36=18x \\ x^2-36=0 \\ x^2=36 \\ x= \sqrt{36} \\$
x₁=6
x₂=-6