# Math Help - Quadratic Equation

I've been working on this equation, here's what I have so far:

$z^2-4z+26=0$
$z^2-4z+4=30$
$(z-2)^2=30$
$z-2= +- Square root of 30$
...

Originally Posted by HellBunny
I've been working on this equation, here's what I have so far:
$z^2-4z+26=0$
$z^2-4z+4=30$
$(z-2)^2=30$
$z-2= \pm \sqrt{30}$
Please use the correct LaTeX tags.
[TEX]z-2= \pm \sqrt{30}[/TEX] gives $z-2= \pm \sqrt{30}$

Originally Posted by HellBunny
I've been working on this equation, here's what I have so far:

$z^2-4z+26=0$
$z^2-4z+4=30$
$(z-2)^2=30$
$z-2= +- Square root of 30$
...
How did you get from the first line to the second? Your quadratic equation has two complex solutions since $b^2-4ac<0$

Okay ( to first poster )

(Second poster)
$[1/2(-4)]^2$
Is how I got the input.

Originally Posted by HellBunny
I've been working on this equation, here's what I have so far:

$z^2-4z+26=0$
$z^2-4z+4=30$
$(z-2)^2=30$
$z-2= +- Square root of 30$
...
This is not correct. It has to be:
$z^2-4z+26=0$
$\Leftrigharrox (z^2-4z+4)+22=0$
$\Leftrightarrow (z-2)^2=-22$
$\Lefrightarrow z-2= \pm \sqrt{-22}$
$\Leftrightarrow z=\pm \sqrt{-22} + 2$

Now, what's $\sqrt{-22}$?

Note that $D<0$ therefore there're only complex roots, like we obtained.

Originally Posted by HellBunny
Okay ( to first poster )

(Second poster)
$[1/2(-4)]^2$
Is how I got the input.
I still don't follow what you did to the value of "c".

If you took 22 from both sides (which is what the LHS suggests) then you'd have -22 on the right (giving complex solutions)
If you added 30 to both sides as implied then you'd have +56 on the left.

Edit: too slow