Originally Posted by
harish21 No.
You are left with $\displaystyle x^2+ 2\times x \times 8 + 6 = 0$.....(I)
So $\displaystyle (x+c)^2 = x^2+ 2.x.c +c^2$...(II)
if you compare I and II, you should see that c = 8.
So you need to have $\displaystyle (x+8)^2$, which is equal to $\displaystyle x^2 + 2.x.8+ 64$
now if you look at equation (I) again, you already have $\displaystyle x^2+ 2\times x \times 8 + 6 = 0$.
To make this $\displaystyle x^2+ 2\times x \times 8 + 64 = 0$,
you need to add 58 right?
so,
$\displaystyle x^2+ 2\times x \times 8 + 6 +58 -58= 0$
Now put the above equation in the form of $\displaystyle (x+c)^2=a$