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How do I solve the following equation?

$x^2 + 10 = 15$

Here's how I think this should be solved. \begin{align*} x^2 + 10 - 10 & = 15 - 10 \\ x^2 & = 15 - 10 \\ x^2 & = 5 \\ x & = \sqrt{5} \end{align*} I was thinking that the square root of 5 is iregular repeating 2.23606797749979 number. 2.236 multipled by itself equals 5ish.

I've also seen another equation like this: \begin{align*} x^2 & = 4 \\ x^2 + 4 & = 0 \\ (x - 2)(x + 2) & = 0 \\ x & = 2 \text{ or } -2 \end{align*} So I guess I could near the end of my equation do the following:

$x^2 + 5 = 0$

and then go from there?

Is my first attempt at solving correct?

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    Going from $x^2=4$ to $x^2+4=0$ is wrong. From $x^2=4$ you can go to $x^2-4=0$.2012-11-28

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Your equation should be corrected at the second line to be $x^2 -4 = 0$ and similarly $x^2-5=0$. Then $(x-\sqrt{5})(x+\sqrt{5})=0$ which implies $x=\sqrt{5}$ or $x=-\sqrt{5}$ The answer being irrational doesn't matter, as $x$ is in the real number set (or can even be complex number set, a quadratic equation always have a solution in the complex number set).

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    Awesome thanks. I had a feeling that I was partly correct.2012-11-28
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The idea is there, however be aware of the fact that $x^2 = a$ is equivalent to $x = \sqrt{a}$ OR $x = -\sqrt{a}$.