В какой точке производная функция у=(х+3)х^2 равна 3?

$y'(x)=[(x+3)*x^2]'=(x+3)'*x^2+(x+3)*(x^2)'=\\\\ =((x)'+(3)')*x^2+(x+3)*2*x^{2-1}=\\\\ =(1+0)*x^2+(x+3)*2x=x^2+2x(x+3)=\\\\ =x^2+2x^2+6x=3x^2+6x\\\\\\ y'(x)=3\\\\ 3x^2+6x=3\\\\ 3x^2+6x-3=0\\\\ 3(x^2+2x-1)=0\\\\ x^2+2x-1=0\\\\ D=2^2-4*1*(-1)=4+4=4*2=(2\sqrt{2})^2\\\\ x_{1,2}=\frac{-2\pm2\sqrt{2}}{2*1}=-1\pm\sqrt{2}$

Ответ: $-1\pm\sqrt{2}$