## Elementary Algebra

We will solve the equation using the property which says that if $x^{2}=a$, then $x=\pm\sqrt a$. Step 1: $(3x-4)^{2}+6=4$ Step 2: $(3x-4)^{2}=4-6$ Step 3: $(3x-4)^{2}=-2$ Step 4: $3x-4=\pm \sqrt {-2}$ Step 5: $3x-4=\pm \sqrt {-1\times2}$ Step 6: $3x-4=\pm (\sqrt {-1}\times\sqrt {2})$ Step 7: $3x-4=\pm (i\times\sqrt {2})$ [as $i=\sqrt {-1}$] Step 8: $3x-4=\pm (i\sqrt {2})$ Step 9: $3x=4\pm (i\sqrt {2})$ Step 10: $x=\frac{4+i\sqrt {2}}{3}$ or $x=\frac{4-i\sqrt {2}}{3}$ Therefore, the solution set is {$\frac{4-i\sqrt {2}}{3},\frac{4+i\sqrt {2}}{3}$}. So, the statement is true and the solution set consists of two, non-real complex numbers.