Question 2913
Completing the square is quite easy. Suppose we have a quadratic equation {{{ f(x) = ax^2 + bx + c = 0}}}. We assume that a is not equal zero, because in that case we couldn't speak of quadratic equation. Now we can write {{{ f(x) = a(x^2 + (b/a)x + (c/a)) = 0 }}} and dividing by a we get {{{ x^2 + (b/a)x + (c/a) = 0 }}}.  As we want to get that formula in something like {{{ (x+m)^2 = n }}} now it's our deal to ensure that {{{ x^2 + (b/a)x + (c/a) = (x+m)^2 - n }}}. This we can do simply by writing {{{ x^2 + (b/a)x + (b/(2a))^2 - (b/(2a))^2 + (c/a) = (x + b/(2a))^2 + (c/a - (b/(2a))^2) }}}. So now we have what we wanted just by putting {{{ m = b/(2a), n = (b/(2a))^2 - c/a }}}. To be more concrete we can show an example of that method. {{{ 2x^2 + 4x + 4 = 0 }}}. First, we divide by 2 an we obtain {{{ x^2 + 2x + 2 = 0 }}}. Now we can write {{{ x^2 + 2x + 2 = x^2 + 2x + 1 - 1 + 2 = (x+1)^2 + (2 - 1) = 0 }}} and so {{{ (x+1)^2 = -1 }}}. From this equation we see that it has no real roots because the square cannot be a negative number.