# How do you solve 6y^2+y+-2=0?

• Interpreting $\pm$ as $+$: the equation becomes $6 {y}^{2} + y + 2 = 0$. The discriminant of $6 {y}^{2} + y + 2$ is negative, and so there are no solutions.
• Interpreting $\pm$ as $-$: the equation becomes $6 {y}^{2} + y - 2 = 0$. The discriminant of $6 {y}^{2} + y - 2$ is positive. So we can solve it, using the formula
${y}_{1 , 2} = \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$
Since $a = 6$, $b = 1$ and $c = - 2$, the formula becomes
${y}_{1 , 2} = \frac{- 1 \setminus \pm \setminus \sqrt{49}}{12} = \frac{- 1 \setminus \pm 7}{12}$
So, ${y}_{1} = \frac{- 1 - 7}{12} = - \frac{8}{12} = - \frac{2}{3}$, and ${y}_{2} = \frac{- 1 + 7}{12} = \frac{6}{12} = \frac{1}{2}$