(i) Given equation is `x-(1)/(x)=3`
`implies(x^(2)-1)/(x)=3`
`impliesx^(2)-1=3ximpliesx^(2)-3x-1=0`
On comparing with `ax^(2)+bx+c=0`
a=1,b=-3 and c=-1
`becausex=(-b+-sqrt(b^(2)-4ac))/(2a)`
`impliesx=(-(3)+-sqrt((-3)^(2)-4xx1xx-1))/(2xx1)`
`impliesx=(3+-sqrt13)/(2)`
`becausex=(3+sqrt13)/(2) and (3-sqrt13)/(2)`
Hence, roots of the equation are `(3+sqrt13)/(2) and (3-sqrt13)/(2)`
(ii) Given equation is `(1)/(x+4)-(1)/(x-7)=(11)/(30)`
`implies((x-7)-(x+4))/((x+4)(x-7))=(11)/(30)implies(-11)/(x^(2)-7x+4x-28)=(11)/(30)`
`implies11(x^(2)-3x-28)=30xx(-11)`
`impliesx^(2)-3x-28=-30`
`impliesx^(2)-3x+2=0`
`impliesx^(2)-(2+1)x+2=0`
`impliesx^(2)-2x-x+2=0`
`impliesx(x-2)-1(x-2)=0`
`implies(x-2)(x-1)=0`
`impliesx=2 and x=1`
Hence, roots of the equation are 2 and 1.