how would you solve
$\displaystyle x^2+|2x-1|=3$
interesting ...
$\displaystyle |2x-1| = 3-x^2$
since $\displaystyle |2x-1| \geq 0$ , $\displaystyle -\sqrt{3} \leq x \leq \sqrt{3}$
case 1 ...
$\displaystyle 2x-1 = 3-x^2$
$\displaystyle x^2+2x-4 = 0$
$\displaystyle x^2+ 2x + 1 - 5 = 0$
$\displaystyle (x + 1)^2 - 5 = 0$
$\displaystyle x+1 = \pm \sqrt{5}$
$\displaystyle x = -1 \pm \sqrt{5}$
one solution works ... $\displaystyle x = -1 + \sqrt{5}$
case 2 ...
$\displaystyle 2x-1 = -(3 - x^2)$
$\displaystyle 2x-1 = x^2 - 3$
$\displaystyle 0 = x^2 - 2x - 2$
$\displaystyle 0 = x^2 - 2x + 1 - 3$
$\displaystyle 0 = (x - 1)^2 - 3$
$\displaystyle x-1 = \pm \sqrt{3}$
$\displaystyle x = 1 \pm \sqrt{3}$
one solution works ... $\displaystyle x = 1 - \sqrt{3}$
recall, $\displaystyle |x| = \left \{ \begin{array}{lr} x & \text{ if } x \ge 0 \\ & \\-x & \text{ if } x < 0 \end{array} \right.$
thus,
$\displaystyle x^2 + |2x - 1| = 3$
$\displaystyle \Rightarrow |2x - 1| = 3 - x^2$
$\displaystyle \Rightarrow 2x - 1 = 3 - x^2$ ...........(if $\displaystyle 2x - 1 \ge 0$)
or
$\displaystyle -(2x - 1) = 3 - x^2$ ...........(if $\displaystyle 2x - 1 < 0$)
now solve each equation independently, and then take the interval that covers both solutions