Answer
The solution set is $\left\{-\frac{2}{3}, \frac{1}{2}\right\}$.
Work Step by Step
Subtract $2$ on both sides of the equation to obtain:
$2-2 = y+6y^2-2
\\0 = 6y^2+y-2$
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:
$x_1 = \dfrac{-1+7}{12} = \dfrac{6}{12} = \dfrac{1}{2}
\\x_2=\dfrac{-1-7}{12}=\dfrac{-8}{12}=-\dfrac{2}{3}$
Therefore, the solution set is $\left\{-\frac{2}{3}, \frac{1}{2}\right\}$.
