Solve 3p^2+8p-28leq0

$$3 p ^ { 2 } + 8 p - 28 \leq 0$$

Solve for p

$$p\in \begin{bmatrix}-\frac{14}{3},2\end{bmatrix}$$

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.

$$3p^{2}+8p-28=0$$

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 -28 for c in the quadratic formula.

$$p=\frac{-8±\sqrt{8^{2}-4\times 3\left(-28\right)}}{2\times 3}$$

Do the calculations.

$$p=\frac{-8±20}{6}$$

Solve the equation p=\frac{-8±20}{6} when ± is plus and when ± is minus.

$$p=2 p=-\frac{14}{3}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(p-2\right)\left(p+\frac{14}{3}\right)\leq 0$$

For the product to be ≤0, one of the values p-2 and p+\frac{14}{3} has to be ≥0 and the other has to be ≤0. Consider the case when p-2\geq 0 and p+\frac{14}{3}\leq 0.

$$p-2\geq 0 p+\frac{14}{3}\leq 0$$

This is false for any p.

$$p\in \emptyset$$

Consider the case when p-2\leq 0 and p+\frac{14}{3}\geq 0.

$$p+\frac{14}{3}\geq 0 p-2\leq 0$$

The solution satisfying both inequalities is p\in \left[-\frac{14}{3},2\right].

$$p\in \begin{bmatrix}-\frac{14}{3},2\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$p\in \begin{bmatrix}-\frac{14}{3},2\end{bmatrix}$$