complex coefficient in quadratic equation question

Question

I got a question I am unable to work out. Since it's quadratic with complex coefficients I believe we should use the quadratic equation but it doesn't seem to work out.

$x^2-(1+i)x+(2+2i)=0$

Below is what I did:

a = 1
b = -(1+i)
c = 2+2i

After plugging in the quadratic equation it is:

$-(-(1+i))\pm\sqrt{({-(1+i))}^2-4(1)(2+2i)}\over2(1)$

$(-1-i)^2 = 1^2+2i+i^2 = 1 + 2i -1 = 2i$

$(1+i)\pm\sqrt{2i-8-8i)}\over2$

$(1+i)\pm\sqrt{-8-6i)}\over2$

I don't think i can simplify it further after that but the answer should be $2i$ and $1-i$.
Did I make a mistake somewhere?

Answer 1

Let's continue from where you stopped. Consider

$$-8-6i=1-9-6i=1^2+(3i)^2-2\times 3i=(3i-1)^2$$

And so the roots are

$$\begin{align} x_1=&{1+i+3i-1\over 2}=2i\\x_2=&{1+i-3i+1\over 2}=1-i\end{align}$$

Answer 2

You can't solve $-(1+i)^2 $ in this type.

$(-1-i)^2 = 1^2+2i+i^2 = 1 + 2i -1 = 2i$

Because on solving,

$-(1+i)^2 $ we get $-2i$

$-(1+i)^2 = -(1^2+2i+i^2) = -(1+2i-1) = -2i$

Then,

$(1+i)\pm\sqrt{-(-2i)-8-8i}\over2$

$(1+i)\pm\sqrt{-8-6i}\over2$

$(1+i)\pm \sqrt{9i^2-6i+1}\over2$

$(1+i)\pm \sqrt{(3i-1)^2}\over2$

$(1+i)\pm (3i-1)\over2$

$ x = \frac{1+i+3i-1}{2} , \frac{1+i-3i+1}{2}$

On solving you get 2i, 1-2i.