Here's the question you clicked on:
neverforgetvivistee
Can you simplify this further? *Using the quadratic formula to factor x^2-6x-4* The answer is supposed to be 6(+-)√53 / 2 ,but can't you simplify by 2? 6 and 2 go into 2. Then could it be 3(+-)√68/1 or no?
6(+-)√53 / 2 is simplified
answer is is \[3\pm\sqrt{13}\]
wait how is it simplified? in another problem there was a problem where you divided by 3 because it was a common factor
wait i'll go look for it
no you cannot because the number inside the radical contains no perfect squares. this is the best you can do
later in the problem the guy got 9(+-)3√33 --------- -18 then he simplified by 3
khanacademy
for \[-9x^2-9x+6=0\] you can start with \[3x^2+3x-2=0\] and reduce before you start
hold the phone solution to \[ x^2-6x-4=0\]is not what you wrote. it is what tomas A wrote
\[\frac{9\pm3\sqrt{33}}{-18}=\] you can simplify this since \[\frac{3(3\pm\sqrt{33})}{-6\cdot3}=\] \[\frac{1(3\pm\sqrt{33})}{-6\cdot1}=\]
\[x^2-6x-4=0\] \[x^2-6x=4\] \[(x-3)^2=4+9=13\] \[x-3=\pm\sqrt{13}\] \[x=3\pm\sqrt{13}\]