What is the value of k if `x=(k+12)/(10)` when the quadratic equation `5x^(2)+2x-7=0` is solved by Sreedhar Acharya's formula ?

Explanation

Comparing the given equation `5x^(2)+2x-7=0` with `ax^(2)+bx+c=0(ane0)` we get, `a=5,b=2andc=-7`
`:.` Accroding to the Sreedhar Acharya's formula, `x=(-bpmsqrt(b^(2)-4ac))/(2a)=(-2pmsqrt((2)^(2)-4xx5xx(-7)))/(2xx5)`
`=(-2pmsqrt(4+140))/(10)=(-2pmsqrt144)/(10)=(-2pm12)/(10)`
`=(-2pm12)/(10)` (taking + sign) and `(-2-12)/(10)` (taking-sign)
Now, given that `x=(k+12)/(2)..........(1).` Also `x=(-2+12)/(10)" ".......(2)`
`:.` Comparing (1) and (2), we get k=-2.
Hence the required value of k=-2.