Solutions of $\sqrt{x^2+5x-14} + |x^2+4x-12|=0$ My attempt:
Given,
$$\sqrt{x^2+5x-14} + |x^2+4x-12|=0 \tag{1}$$
Since $|a|=\sqrt{a^2}$, $$\sqrt{x^2+5x-14}=-\sqrt{(x^2+4x-12)^2}$$
Squaring both sides, $$x^2+5x-14=(x^2+4x-12)^2$$
When I simplify the above, I get two real solutions: $x=2$ and $x=2.138 \text{ (approximately)}$. However, there is only one solution to equation $(1)$ according to Wolfram Alpha, $x=2$.
Is my solution incorrect? If so, where did I go wrong?

Why not factorise first, since both quadratics clearly very easily factorise, into $\sqrt{(x+7)(x-2)} + |(x+6)(x-2)| = 0 \implies$
$\sqrt{(x+7)(x-2)} = -|(x+6)(x-2)| \implies $
$(x+7)(x-2) = ((x+6)(x-2))^2$
This very clearly only has one solution, being $x=2$, since both terms must be zero
