Решите уравнение (√2(x²-6x)+19)+(√3x²-18x+36)=-2x²+12x-14

Решите задачу:

$\sqrt{2(x^2-6x)+19}+\sqrt{3(x^2-6x)+36} = -2(x^2-6x)-14\\ 3x^2-18x+36=3(x^2-6x+12)\\ x^2-6x+12=t\\ \sqrt{2t-5}+\sqrt{3t}=-2t+10\\ 2t-5+3t+2\sqrt{6t^2-15t}=10-2t\\ \sqrt{6t^2-15t}*2=15-7t\\ (6t^2-15t)*4=225-210t+49t^2 24t^2-60t=225-210t+49t^2\\ 25t^2-150t+225=0\\ t^2-6t+9=0 \\ (t-3)^2=0\\ t=3\\ x^2-6x+12=3\\ x^2-6x+9=0\\ (x-3)^2=0\\ x=3$