Let `f(x)=3x^(2)-5x-1`. Then solve f(x)=0 by
(i) factroing the quadratic
(ii) using th quadratic formula
(iii) completing the square
and then rewrite f(x) in the form `A(x+-B)^(2)+-C`.

Explanation

`f(x)=3x^(2)-5x-1`
:.f(x)=0
`implies3x^(2)-5x-1=0`
(i) The given quadratic equation cannot be fully factorised using real integers. So it is better to solve this equation by any other method.
(ii) `3x^(2)-5x-1=0`
Compare it with `ax^(2)+bx+c=0`, we get
a=3, b=-5, c=-1
:. Let two roots of this equation are
`alpha=(-b+sqrt(b^(2)-4ac))/(2a)andbeta=(-b-sqrt(b^(2)-4ac))/(2a)`
`(-(5)+sqrt((-5)^(2)-4(3)(-1)))/(2xx3)`
`=(5+sqrt37)/(6)`
`=(-(-5)-sqrt((-5)^(2)-4(3)(-1)))/(2xx3)`
`beta=(5-sqrt(37))/(6)`
:. Two values of x are `(5+sqrt(37))/(6)and(5-sqrt37)/(6)`
(iii) `3x^(2)-5x-1=0`
`impliesx^(2)-(5)/(3)x(1)/(3)=0" "("dividing both sides by 3")`
`impliesx^(2)-(5)/(3)x+.....=(1)/(3)+.....`
`impliesx^(2)-(5)/(3)x+((5)/(6))^(2)=(1)/(3)+((5)/(6))^(2)" "["adding"(("coeff. of x")^(2)/(2))"on both sides"]`
`implies(x-(5)/(6))^(2)=(1)/(3)+(25)/(36)`
`implies(x-(5)/(6))^(2)=(12+25)/(36)`
`implies(x-(5)/(6))^(2)=((sqrt37)/(6))^(2)`
`impliesx-(5)/(6)=+-(sqrt37)/(6)`
`:.x=(5)/(6)+(sqrt37)/(6)=(5+sqrt37)/(6),x=(5)/(6)-(sqrt37)/(6)=(5-sqrt37)/(6)`
:. Two values of x are `(5+sqrt37)/(6)and(5-sqrt(37))/(6)`.
Now, `f(x)=3x^(2)-5x-1=3(x^(2)-(5)/(3)x)-1`
`=3(x^(2)-(5)/(3)x+(25)/(36)-(25)/(36))-1`
`=3(x-(5)/(6))^(2)-(25)/(12)-1=3(x-(5)/(6))^(2)-(37)/(12)`
which is of the form `A(x-B)^(2)-C`, where A=3, `B=(5)/(6),C=(37)/(12)`.