Найдите корень уравнения √1-x=x+1 помогите плз

Решите задачу:

$\sqrt{1-x}=x+1 \\ \\ \left \{ {{1-x=(x+1)^2} \atop {1-x \geq 0,x+1 \geq 0}} \right. \\ \\ \left \{ {{1-x=x^2+2x+1} \atop {x \leq 1,x \geq -1}} \right. \\ \\ \left \{ {{x^2+3x=0} \atop {x E[-1; 1]}} \right. \\ \\ \left \{ {{x*(x+3)=0} \atop {x E[-1; 1]}} \right. \\ \\ \left \{ {{x=0, x=-3} \atop {x E[-1; 1]}} \right. \\ x=0$