Solve 6x^2+33x+36leq0

$$6 x ^ { 2 } + 33 x + 36 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-4,-\frac{3}{2}\end{bmatrix}$$

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.

$$6x^{2}+33x+36=0$$

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 6 for a, 33 for b, and 36 for c in the quadratic formula.

$$x=\frac{-33±\sqrt{33^{2}-4\times 6\times 36}}{2\times 6}$$

Do the calculations.

$$x=\frac{-33±15}{12}$$

Solve the equation x=\frac{-33±15}{12} when ± is plus and when ± is minus.

$$x=-\frac{3}{2} x=-4$$

Rewrite the inequality by using the obtained solutions.

$$6\left(x+\frac{3}{2}\right)\left(x+4\right)\leq 0$$

For the product to be ≤0, one of the values x+\frac{3}{2} and x+4 has to be ≥0 and the other has to be ≤0. Consider the case when x+\frac{3}{2}\geq 0 and x+4\leq 0.

$$x+\frac{3}{2}\geq 0 x+4\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+\frac{3}{2}\leq 0 and x+4\geq 0.

$$x+4\geq 0 x+\frac{3}{2}\leq 0$$

The solution satisfying both inequalities is x\in \left[-4,-\frac{3}{2}\right].

$$x\in \begin{bmatrix}-4,-\frac{3}{2}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-4,-\frac{3}{2}\end{bmatrix}$$