Write 2x + x^2 - 1 in standard form and solve 2x + x^2 - 1 = 0.

Step-by-step solution

1. 1

   Reorder into standard form. Standard form lists terms by descending degree, which is what every quadratic method assumes:

   $2x+x^{2}-1=x^{2}+2x-1$

   so $a=1$, $b=2$, $c=-1$. Reordering changes nothing about the value — addition is commutative — but it prevents misreading $2$ as the leading coefficient.

2. 2

   Check for integer factoring, then discard it. Two integers with product $-1$ and sum $2$ would be needed; the only integer pair for $-1$ is $(1,-1)$, whose sum is $0$. So no integer factorisation exists and the roots are irrational.

3. 3

   Complete the square — quicker than the formula when $b$ is even. Half of $b=2$ is $1$, so

   $x^{2}+2x-1=(x+1)^{2}-1-1=(x+1)^{2}-2$

   Setting this to zero gives $(x+1)^{2}=2$, a form that can be solved by inspection.

4. 4

   Take square roots, keeping both signs.

   $x+1=\pm\sqrt{2}\qquad\Longrightarrow\qquad x=-1\pm\sqrt{2}$

5. 5

   Cross-check with the quadratic formula.

   $x=\frac{-2\pm\sqrt{4+4}}{2}=\frac{-2\pm2\sqrt{2}}{2}=-1\pm\sqrt{2}\ \checkmark$

   (Using $\sqrt{8}=2\sqrt2$ so the $2$ cancels.) Vieta confirms it too: the roots sum to $-2=-b/a$ ✓ and multiply to $1-2=-1=c/a$ ✓. Numerically $x\approx0.4142$ and $x\approx-2.4142$, residuals below $10^{-12}$ ✓.