Solve y^2-y-6<0

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

Solve for y

$$y\in \left(-2,3\right)$$

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.

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

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 1 for a, -1 for b, and -6 for c in the quadratic formula.

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

Do the calculations.

$$y=\frac{1±5}{2}$$

Solve the equation y=\frac{1±5}{2} when ± is plus and when ± is minus.

$$y=3 y=-2$$

Rewrite the inequality by using the obtained solutions.

$$\left(y-3\right)\left(y+2\right)<0$$

For the product to be negative, y-3 and y+2 have to be of the opposite signs. Consider the case when y-3 is positive and y+2 is negative.

$$y-3>0 y+2<0$$

This is false for any y.

$$y\in \emptyset$$

Consider the case when y+2 is positive and y-3 is negative.

$$y+2>0 y-3<0$$

The solution satisfying both inequalities is y\in \left(-2,3\right).

$$y\in \left(-2,3\right)$$

The final solution is the union of the obtained solutions.

$$y\in \left(-2,3\right)$$