Solve 2y^2-y-6>0

$$2 y ^ { 2 } - y - 6 > 0$$

Solve for y

$$y\in \left(-\infty,-\frac{3}{2}\right)\cup \left(2,\infty\right)$$

Steps Using the Quadratic Formula

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$2y^{2}-y-6=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 2 for a, -1 for b, and -6 for c in the quadratic formula.

$$y=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Do the calculations.

$$y=\frac{1±7}{4}$$

Solve the equation y=\frac{1±7}{4} when ± is plus and when ± is minus.

$$y=2 y=-\frac{3}{2}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(y-2\right)\left(y+\frac{3}{2}\right)>0$$

For the product to be positive, y-2 and y+\frac{3}{2} have to be both negative or both positive. Consider the case when y-2 and y+\frac{3}{2} are both negative.

$$y-2<0 y+\frac{3}{2}<0$$

The solution satisfying both inequalities is y<-\frac{3}{2}.

$$y<-\frac{3}{2}$$

Consider the case when y-2 and y+\frac{3}{2} are both positive.

$$y+\frac{3}{2}>0 y-2>0$$

The solution satisfying both inequalities is y>2.

$$y>2$$

The final solution is the union of the obtained solutions.

$$y<-\frac{3}{2}\text{; }y>2$$