Solve y^2-y-6leq0

$$y ^ { 2 } - y - 6 \leq 0$$

Solve for y

$$y\in \begin{bmatrix}-2,3\end{bmatrix}$$

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.

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

For the product to be ≤0, one of the values y-3 and y+2 has to be ≥0 and the other has to be ≤0. Consider the case when y-3\geq 0 and y+2\leq 0.

$$y-3\geq 0 y+2\leq 0$$

This is false for any y.

$$y\in \emptyset$$

Consider the case when y-3\leq 0 and y+2\geq 0.

$$y+2\geq 0 y-3\leq 0$$

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

$$y\in \begin{bmatrix}-2,3\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$y\in \begin{bmatrix}-2,3\end{bmatrix}$$