Решите уравнения через дискриминант

Решите задачу:

$1)\; \; x^2+2(1+\sqrt8)x+8\sqrt2=0\\\\\frac{D}{4}=(\frac{b}{2})^2-ac=(1+\sqrt8)^2-2\sqrt8= 9+2\sqrt8-8\sqrt2=9\\\\x_{1,2}=\frac{-\frac{b}{2}\pm \sqrt{\frac{D}{4}}}{a}=-(1+\sqrt8)\pm 3\\\\x_1=-4-\sqrt8=-4-2\sqrt2=-2(2+\sqrt2)\; ,\\\\x_2=2-\sqrt8=2-2\sqrt2=2(1-\sqrt2)\\\\2)\; \; x^2+4x-\sqrt3+1=0\\\\\frac{D}{4}=2^2-(1-\sqrt3)=3+\sqrt3\\\\x_{1,2}=-2\pm \sqrt{3+\sqrt3}\\\\x_1=-2-\sqrt{3+\sqrt3}\\\\x_2=-2+\sqrt{3+\sqrt3}$

$3)\; \; \frac{(x-1)^2}{5}-\frac{x+4}{6}=\frac{2x-2}{3}\; |\cdot 30\\\\6(x^2-2x+1)-5(x+4)=10(2x-2)\\\\6x^2-12x+6-5x-20-20x+20=0\\\\6x^2-37x+6=0\\\\D=b^2-4ac=1225=35^2\; ,\; \; \; x_{1,2}=\frac{37\pm 35}{12}\\\\x_1=\frac{2}{12}=\frac{1}{6}\; \; ,\; \; x_2=\frac{72}{12}=6$