在直角坐標平面上一條連接(2x-1,2x+1)和(x-6,1-3x)兩點的線段長度是17。求x的可能性?

[(2x-1)-(x-6)]^2+[(2x+1)-(1-3x)]^2=17^2
(x+5)^2+(5x)^2=289
x^2+10x+25+25x^2=289
26x^2+10x-264=0
x=3 or x=-44/13

17^2=[(2x-1)-(x-6)]^2+[(2x+1)-(1-3x)]^2
289=(x+5)^2+(5x)^2
289=26x^2+10x+25
26x^2+10x-264=0
13x^2+5x-132=0
(13x+44)(x-3)=0
x= - 44/13 or 3