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\[(-b \pm \sqrt{b ^{2}-4ac)}/2a\]

b=-8, a=1, and c=8

I prefer:
\[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
\.[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]

that's more betterr

6.828, 1.17157

ok so if i have \[8\pm 4\sqrt{2}/2\]..how would i solve that?

You could simplify further and get
\[4\pm 2\sqrt{2}\]

Solve for both roots

Use the + value for one root, then use the - for the other
\[4-2\sqrt{2}\]

\[4+2\sqrt{2}\]

soo would it be 2 sqrt 2 for one and 6 sqrt 2 for the other

Let me know what you get.

He has solved it down to the point of\[4\pm 2\sqrt{2}\]

radar ..when u get done will you help me?

or chatindia..lol

Yes, would be glad too, providing I can. I am just reviewing math from the past lol

lol okay thank you! :))

Is your problem already posted laurenlynn?

yes

it's the one that says PLEASE HELP ME!!! at the top..haha

O.K. I will look for it.