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This is a simple homework thing for my college. Is there a way to isolate $y$ in this case: $$ y ^ 2 + y = x $$ I tried everything and I can't solve this. Thanks.

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    $\begingroup$ Quadratic formula or completing the square would be a couple of techniques I'd suggest as either may work. $\endgroup$ – JB King Jan 11 '14 at 0:19
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    $\begingroup$ try this $y^2 + y = (y+ 1/2)^2 - 1/4$ $\endgroup$ – math student Jan 11 '14 at 0:20
  • $\begingroup$ @LeandroTavares That doesn't make sense to me. How did you arrive to that formula? Sorry if I am a beginner, still learning math at college. $\endgroup$ – Pacha Jan 11 '14 at 0:23
  • $\begingroup$ @JBKing Tried them, doesn't work. $\endgroup$ – Pacha Jan 11 '14 at 0:23
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    $\begingroup$ Pacha: @LeandroTavares' method is one of the standard tricks they teach you in pre-algebra: it's called "completing the square". $\endgroup$ – Hurkyl Jan 11 '14 at 1:02
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$y^2 + y = x => (y + 1/2)^2 - 1/4 = x => (y+ 1/2)^2 = x + 1/4 $

Then $(y + 1/2) = \pm \sqrt{x+ 1/4}$ which implies $y = \pm \sqrt{x+ 1/4} - (1/2)$

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    $\begingroup$ The last $+$ should be a $-$. $\endgroup$ – Pedro Tamaroff Jan 11 '14 at 2:14
  • $\begingroup$ @PedroTamaroff thank you, i fixed the error $\endgroup$ – math student Jan 11 '14 at 21:14
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Initial equation:

$y^2+y-x=0$

Applying Quadratic formula in terms of y, $a=1$, $b=1$, $c=-x$:

$y = \frac{-1\pm\sqrt{1+4x}}{2}$

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  • $\begingroup$ It may be helpful to realize that when you apply the quadratic formula, $x$ should be thought of as a constant. The equation $y^2+y-x=0$ is of the form $ay^2 +by+c=0$, by setting $a=1$, $b=1$, and $c=-x$. If this is too complicated, first try solving e.g. $y^2 + y = 3$ and $y^2 + y = 5$. Once you get how to solve these two, you can apply the technique to the general case $y^2 + y = x$. $\endgroup$ – yyzz Jan 11 '14 at 0:41

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