Simplify algebraic expression

This is basic algebra. I must be making a mistake somewhere. Where is it?

I start out with $x^2 - x = y - 1$.

$x^2 - x - (y - 1) = 0$.

Using quadratic formula, $x = \frac{x \pm \sqrt{x^2 + 4(y - 1)}}{2}$.

$x = \pm\sqrt{x^2 + 4(y - 1)}$

$x^2 = x^2 + 4(y-1)$.

$0 = 4(y -1)$

$0 = y - 1$

$y = 1$

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What exactly are you trying to do? If you are trying to solve for x in terms of the coefficients and $y$ (which I assume is being held constant) then you only take the coefficient of the $x$ terms when you apply the quadratic formula, so it would be $$x=\frac{1\pm\sqrt{1+4(y-1)}}{2} \\x=\frac{1\pm\sqrt{4y-3}}{2}$$ which cannot be simplified any further.

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Yes, thank you, I put x and x^2 instead of their coefficients. Silly misapplication of quadratic formula. –  user75122 Jun 9 '13 at 2:26

$x=\dfrac{1 \pm \sqrt{1+4(y-1)}}{2}$,don't put $x$ in.

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Yep, this question is resolved, thanks everyone –  user75122 Jun 9 '13 at 2:28

If you mean for $y$ to be a constant, then $$x^2 - x = y - 1 \iff x^2 + -x + -(y-1)$$ So if you want to use the quadratic formula to "solve" for x:

$$x = \dfrac {-b\pm \sqrt{b^2 - 4ac}}{2a}$$

then you need to substitute $a = 1, b = -1, c = -(y-1)$: $$x=\dfrac{1 \pm \sqrt{1+4(y-1)}}{2}$$No $x$ terms in the right-hand side of the quadratic formula when solving for $x$!

Otherwise, I will assume you have two variables: $x, y$, which is not quadratic equation in one variable, so the quadratic formula is inappropriate. So I will assume, as your title suggests, that you might have been asked to simplify the equation, or to express it as a function of $y$, or to express it in a standard form for a parabola:

You can take your equation and write it as a function of $y$:

$$x^2 - x = y - 1$$ $$y = x^2 - x + 1$$

The last equation reveals nicely that your function $f(x) = y$ is a parabola.

You can "complete the square" to write the parabola in the following form: \begin{align} y & = x^2 - x + 1 \\ & = (x^2 - x - 1/4) + 5/4 \\ & = (x - 1/2)^2 + 5/4 \end{align}

This gives a parabola in the form $$\left(y - \dfrac 54\right) = \left(x - \dfrac 12\right)^2$$ which opens "up", and has its vertex at $\left( \dfrac 12, \dfrac 54\right)$:

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