Solve 5x^2-7x+2=0 completing the square. what is the constant added on to form the perfect square trinomial

Best to factor 5 out of the first 2 terms:

5x^2-7x+2=0  =>  5(x^2 - [7/5]x)                          +2=0

Take half of [-7/5]:  That'd be -7/10.

Square 7/10, add it to  5(x^2 - [7/5]x)                          +2=0  and then subtract it:

5(x^2 - [7/5]x)+ 49/100       -  49/100       )                  +2=0

Then we have   5(x- [7/10] )^2 - 49/100 ) + 200/100 = 0

                          5(x-7/5)^2  - 245/100 + 200/100 = 0
                           5(x-7/5)^2 - 45/100 = 0
Dividing all terms by 5,  x-7/5 = plus or minus sqrt( 9/100 )
                                       x-7/5 = plus or minus 3/10

Then one root is x = 7/5 + 3/10, or  x= 17/10  (answer #1)

The other is        x = 7/5 - 3/10, or x = 14/10 - 3/10, or x = 7/10  (answer #2) 

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