\\\\ \sqrt{x^2+7}=\sqrt{3-4x}\ |^2\\ x^2+7=3-4x\\ x^2+4x+4=0\\ (x+2)^2=0\\ x=-2\\ " alt=" \\\sqrt{x^2+7}=\sqrt{3-4x}\\\\ x^2+7\geq0\\ x^2\geq-7\\ x\in\mathbb{R}\\ 3-4x\geq0\\ -4x\geq-3\\ x\leq\frac{3}{4}\\ D=(-\infty,\frac{3}{4}>\\\\ \sqrt{x^2+7}=\sqrt{3-4x}\ |^2\\ x^2+7=3-4x\\ x^2+4x+4=0\\ (x+2)^2=0\\ x=-2\\ " align="absmiddle" class="latex-formula">