Answer
The solution set is $\left\{-\frac{1}{2}, \frac{2}{3}\right\}$.
Work Step by Step
Subtract $2$ and $z$ on both sides of the equation to obtain:
$2+z-2-z = 6z^2 -2-z
\\0 = 6z^2-z-2$
RECALL:
A quadratic equation of the form $ax^2+bx+c=0$ can be solved using the quadratic formula
$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$
The given quadratic equation has:
$a=6
\\b=-1
\\c=-2$
Substitute these values into the quadratic formula to obtain:
$z=\dfrac{-(-1) \pm \sqrt{(-1)^2-4(6)(-2)}}{2(6)}
\\z=\dfrac{1 \pm \sqrt{1-(-48)}}{12}
\\z=\dfrac{1\pm \sqrt{1+48}}{12}
\\x =\dfrac{1\pm\sqrt{49}}{12}
\\x=\dfrac{1\pm\sqrt{7^2}}{12}
\\x=\dfrac{1\pm 7}{12}$
Split the solutions to obtain:
$x_1 = \dfrac{1+7}{12} = \dfrac{8}{12} = \dfrac{2}{3}
\\x_2=\dfrac{1-7}{12}=\dfrac{-6}{12}=-\dfrac{1}{2}$
Therefore, the solution set is $\left\{-\frac{1}{2}, \frac{2}{3}\right\}$.
