I got this question but I'm not really sure how one part of the solutions came up

$\displaystyle tx^2+4tx+10=t $

Make x the subject in each of the following and give the values of t for which real solution(s) to the equation can be found.

So I complete the square and get up to

$\displaystyle tx^2+4tx+10=t $

$\displaystyle \frac {tx^2+4tx+10-t}{t}=0$

$\displaystyle x^2+4x=\frac{-10}{t} +1 $ I moved $\displaystyle \frac{10}{t}-1 $ to the other side

$\displaystyle (x+2)^2 = \frac {-10}{t} +5 $

$\displaystyle (x+2)^2 = \frac {5(-2+t)}{t} $

So I just take the square root of both sides and subtract -2.

But the solutions say $\displaystyle x=-2 \pm\frac {\sqrt{5t(-2+t)}}{t} $

How come my answer is different...