If the roots of the equation `ax^2+bx+c=0(a!=0)` be equal then

Explanation

Let `5x^(2) - 6x = y`, Then,
`sqrt(5x^(2)-6x+8)-sqrt(5x^(2)-6x-7)=1`
or ` sqrt(y+8)-sqrt(y-7)=1`
or `(sqrt(y+8)-sqrt(y-7))^(2) = 1`
or ` y = sqrt(y^(2)+y-56)`
or `y^(2) = Y^(2) + y- 56`
or y = 56
`rArr 5x^(2)-6x = 56 [because y = 5x^(2) - 6x`]
or (5x+ 14) (x - 4) = 0
or `x = 4,(-14)/(5)`
Clearly, both the values satisfy the given equation. Hence, the roots of the given equation are 4 and `-14//5`.