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$$x^2 + 7x = 2$$

I came across a question similar to this one on my GED book, but instead of $= 2$ it was $= 0$. I changed the number to try and understand it better, but it just got much more difficult.So what are the steps to find out what $x$ is equal to? I have no idea

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    $\begingroup$ Quadratic Formula. Or completing the square. Those could also have been used on $x^2+7x=0$, but for that one there is a much easier way. $\endgroup$ – André Nicolas Apr 8 '14 at 0:19
  • $\begingroup$ Thanks for the answer but how do the steps for Quadratic Formula look like? They don't go over that on the GED book. $\endgroup$ – Mark Apr 8 '14 at 0:39
  • $\begingroup$ If $a\ne 0$, then the solutions of $ax^2+bx+c=0$ are $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. In our case, we are looking at $x^2+7x-2=0$. The solutions are $\frac{-7\pm\sqrt{57}}{2}$. Perhaps the Quadratic Formula is not part of the GED curriculum, I don't know. $\endgroup$ – André Nicolas Apr 8 '14 at 0:44
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Hint: Form a square on the left hand side by adding what's missing (add the same thing on the right hand side too in order to preserve the equality): $$ x^2 +7x +\frac{49}{4} = 2+ \frac{49}{4}.$$ Then take square roots on both sides.

Remember that $(a+b)^2 = a^2+2ab +b^2$.

Then remember that $\sqrt{c^2}=|c|$.

Finally, remember that, if $|c| = d$, where $d\geq 0$, then $c = d$ or $c=-d$.

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