## Elementary Algebra

Using Property 10.1, which states that for any non-negative real number $a$, $x^{2}=a$ can be written as $x=\pm\sqrt a$, we obtain: Step 1: $(2x+9)^{2}=-6$ Step 2: $2x+9=\pm \sqrt {-6}$ Since $\sqrt {-6}$ does not have any real number solutions, $x$ does not have any real number solutions either.