Solve 3x^2-8x-60leq0

$$3 x ^ { 2 } - 8 x - 60 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-\frac{10}{3},6\end{bmatrix}$$

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.

$$3x^{2}-8x-60=0$$

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 3 for a, -8 for b, and -60 for c in the quadratic formula.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\left(-60\right)}}{2\times 3}$$

Do the calculations.

$$x=\frac{8±28}{6}$$

Solve the equation x=\frac{8±28}{6} when ± is plus and when ± is minus.

$$x=6 x=-\frac{10}{3}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-6\right)\left(x+\frac{10}{3}\right)\leq 0$$

For the product to be ≤0, one of the values x-6 and x+\frac{10}{3} has to be ≥0 and the other has to be ≤0. Consider the case when x-6\geq 0 and x+\frac{10}{3}\leq 0.

$$x-6\geq 0 x+\frac{10}{3}\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-6\leq 0 and x+\frac{10}{3}\geq 0.

$$x+\frac{10}{3}\geq 0 x-6\leq 0$$

The solution satisfying both inequalities is x\in \left[-\frac{10}{3},6\right].

$$x\in \begin{bmatrix}-\frac{10}{3},6\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{10}{3},6\end{bmatrix}$$