How do you solve $6 y^{2} + y \pm 2 = 0$ $6y^2+y+-2=0$ ?

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Answer 1 · 2015-03-27T19:26:59

You simply need to solve the two equations: one with the plus sign, and the other with the minus sign.

Interpreting $\pm$ $pm$ as $+$ $+$ : the equation becomes $6 y^{2} + y + 2 = 0$ $6y^2+y+2=0$ . The discriminant of $6 y^{2} + y + 2$ $6y^2+y+2$ is negative, and so there are no solutions.
Interpreting $\pm$ $pm$ as $-$ $-$ : the equation becomes $6 y^{2} + y - 2 = 0$ $6y^2+y-2=0$ . The discriminant of $6 y^{2} + y - 2$ $6y^2+y-2$ is positive. So we can solve it, using the formula
$y_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $y_{1,2}={-b\pm\sqrt(b^2-4ac)}/{2a}$
Since $a = 6$ $a=6$ , $b = 1$ $b=1$ and $c = - 2$ $c=-2$ , the formula becomes
$y_{1, 2} = \frac{- 1 \pm \sqrt{49}}{12} = \frac{- 1 \pm 7}{12}$ $y_{1,2}={-1\pm\sqrt(49)}/{12}={-1\pm 7}/{12}$
So, $y_{1} = \frac{- 1 - 7}{12} = - \frac{8}{12} = - \frac{2}{3}$ $y_1={-1-7}/{12}=-8/12=-2/3$ , and $y_{2} = \frac{- 1 + 7}{12} = \frac{6}{12} = \frac{1}{2}$ $y_2={-1+7}/{12}=6/12=1/2$

How do you solve 6y^2+y+-2=06y2+y±2=06y^2+y+-2=0?