Умоляю, срочно,пожалуйста!

Решите задачу:

$1)2x^{2}-4\sqrt{2}x+3=0\\\\D=(-4\sqrt{2} )^{2}-4*2*3=32-24=8=(2\sqrt{2} )^{2} \\\\x_{1}=\frac{4\sqrt{2}+2\sqrt{2} }{4}=\frac{6\sqrt{2} }{4}=\frac{3\sqrt{2} }{2} \\\\x_{2}=\frac{4\sqrt{2} -2\sqrt{2} }{4}=\frac{2\sqrt{2} }{4}=\frac{\sqrt{2} }{2}$

$2)x^{2} -x(\sqrt{7}-2)-2\sqrt{7}=0\\\\D=(\sqrt{7}-2)^{2}-4*(-2\sqrt{7})=7-4\sqrt{7}+4+8\sqrt{7}=11+4\sqrt{7}=(\sqrt{7}+2)^{2}\\\\x_{1} =\frac{\sqrt{7}-2+\sqrt{7}+2}{2}=\frac{2\sqrt{7} }{2}=\sqrt{7}\\\\x_{2}=\frac{\sqrt{7}-2-\sqrt{7}-2}{2} =\frac{-4}{2}=-2$

$3)2x^{2} -4x-17=0\\\\D=(-4)^{2} -4*2*(-17)=16+136=152=(2\sqrt{38})^{2} \\\\x_{1}=\frac{4+2\sqrt{38} }{4}=\frac{2+\sqrt{38} }{2}\\\\x_{2}=\frac{4-2\sqrt{38} }{4} =\frac{2-\sqrt{38} }{2}$