Need help factoring quadratic terms..5x2-2x+4=0 by completing the square. Really having trouble on this one.?

Question

So far I'm down to..

x2-2/5x+4/5=0

x2-2/5x+2/5=-2/5

Not sure if i'm going the right way thou. Would appreciate any help.

Answer 1

x^2 - (2/5)x + 4/5 = 0
x^2 - (2/5)x + (1/5)^2 + 4/5 - (1/5)^2 = 0
(x - 1/5)^2 + (20 - 1)/25 = 0
(x - 1/5)^2 + 19/25 = 0
(x - 1/5 + (j/5)√19)(x - 1/5 - (j/5)√19) = 0
x = - (1/5)(1 - j√19), - (1/5)(1 + j√19)

Answer 2

its like this.
transpose the constant to the other side:
in this case:
x^2 -2/5x=-4/5
then:
get the half of the numerical coefficient of x, square it and add it to both sides.
so the next step will be:

x^2 -2/5 x+ 1/25 = -4/5 + 1/25
factor the terms on the left side of the equation:
combine like terms on the right side of the equation

so it will be lyk this:
(x-1/5)^2= -19/25
then get the square root of both sides then write the positive egative signs at the right side:

x- 1/5= +-square root of -19/25

evaluate:
x-1/5= + square root of -19/ 5 (since 25 is a perfect square you can get its root)

since there is a negative sign in 19 use the complex number i to remove it. transpose -1/5 to the other side

so it will be:
x= i* square root of 19 /5 +1/5
combine like terms
x= (i*square root of 19 +1) /5

on the case of the negative part of the equation wherein:
x-1/5 =-square root of -19/ 5
evaluate:
x=(-i* square root of 19 + 1) / 5