Помогите решить уравнение. Подробно.

Решите задачу:

$1)\; \; 3x+1+\sqrt{7-9x}=0\\\\\sqrt{7-9x}=-(3x+1)\; \; \to \; \; ODZ:\; \; \left \{ {{3x+1 \leq 0} \atop {7-9x \geq 0}} \right. \; \left \{ {{x \leq -\frac{1}{3}} \atop {x \leq \frac{7}{9}}} \right. \; ,\; x \leq -\frac{1}{3}\\\\\\\\7-9x=(-1)^2\cdot (3x+1)^2\\\\7-9x=9x^2+6x+1\\\\9x^2+15x-6=0\\\\3x^2+5x-2=0\\\\D=25+24=49\; ,\\\\x_1=\frac{-5-7}{6}=-2\; ;\; \; x_2=\frac{-5+7}{6}=\frac{1}{3}\notin ODZ\\\\Otvet:\; \; x=-2\; .$

$2)\quad 1+2x+\sqrt{7-6x}=0\\\\\sqrt{7-6x}=-(1+2x)\; ,\; \; \; ODZ:\; 1+2x \leq 0\; ,\; \; x \leq -\frac{1}{2}\\\\7-6x=1+4x+4x^2\\\\4x^2+10x-6=0\\\\2x^2+5x-3=0\\\\D=25+2\cdot 2\cdot 3=49\\\\x_1=\frac{-5-7}{4}=-3\; ;\; \; x_2=\frac{-5+7}{4}=\frac{1}{2}\notin ODZ$

$Otvet:\; \; x=-3\; .$