Solve 3x^2+14x-5leq0

$$3 x ^ { 2 } + 14 x - 5 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-5,\frac{1}{3}\end{bmatrix}$$

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.

$$3x^{2}+14x-5=0$$

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, 14 for b, and -5 for c in the quadratic formula.

$$x=\frac{-14±\sqrt{14^{2}-4\times 3\left(-5\right)}}{2\times 3}$$

Do the calculations.

$$x=\frac{-14±16}{6}$$

Solve the equation x=\frac{-14±16}{6} when ± is plus and when ± is minus.

$$x=\frac{1}{3} x=-5$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-\frac{1}{3}\right)\left(x+5\right)\leq 0$$

For the product to be ≤0, one of the values x-\frac{1}{3} and x+5 has to be ≥0 and the other has to be ≤0. Consider the case when x-\frac{1}{3}\geq 0 and x+5\leq 0.

$$x-\frac{1}{3}\geq 0 x+5\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{1}{3}\leq 0 and x+5\geq 0.

$$x+5\geq 0 x-\frac{1}{3}\leq 0$$

The solution satisfying both inequalities is x\in \left[-5,\frac{1}{3}\right].

$$x\in \begin{bmatrix}-5,\frac{1}{3}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-5,\frac{1}{3}\end{bmatrix}$$