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\[ x = {-b \pm \sqrt{b^2 - 4ac} \over 2a}\]Hinty hint.
is my answer correct?
x = -2 +/- 1sqroot 13
No... :/
is it okay if you can show me the correct way?
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\[x = {-b \pm \sqrt{b^2 - 4ac} \over 2a}\]You already know what a, b and c are... don't you?
yes a = 1 b = 4 c = -9
Yeah, but you plug the values into the formula that I gave you.
??
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You can simply do that one yourself. \[y = x + 3 \]Suppose they ask you what \(y\) is when \(x = 6\). You plug-in \(x = 6\).\[\implies y = 6 + 3\\\implies y=9 \]
Does that example help?
ok
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