Would it be (*x* + 6)2 = 58 ?

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 \(\displaystyle (x+c)^2=a\)