Solve -12x^2-36x+31geq0

$$- 12 x ^ { 2 } - 36 x + 31 \geq 0$$

Solve for x

$$x\in \begin{bmatrix}-\frac{\sqrt{174}}{6}-\frac{3}{2},\frac{\sqrt{174}}{6}-\frac{3}{2}\end{bmatrix}$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -12x^{2}-36x+31 positive. Since -1 is negative, the inequality direction is changed.

$$12x^{2}+36x-31\leq 0$$

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.

$$12x^{2}+36x-31=0$$

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 12 for a, 36 for b, and -31 for c in the quadratic formula.

$$x=\frac{-36±\sqrt{36^{2}-4\times 12\left(-31\right)}}{2\times 12}$$

Do the calculations.

$$x=\frac{-36±4\sqrt{174}}{24}$$

Solve the equation x=\frac{-36±4\sqrt{174}}{24} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{174}}{6}-\frac{3}{2} x=-\frac{\sqrt{174}}{6}-\frac{3}{2}$$

Rewrite the inequality by using the obtained solutions.

$$12\left(x-\left(\frac{\sqrt{174}}{6}-\frac{3}{2}\right)\right)\left(x-\left(-\frac{\sqrt{174}}{6}-\frac{3}{2}\right)\right)\leq 0$$

For the product to be ≤0, one of the values x-\left(\frac{\sqrt{174}}{6}-\frac{3}{2}\right) and x-\left(-\frac{\sqrt{174}}{6}-\frac{3}{2}\right) has to be ≥0 and the other has to be ≤0. Consider the case when x-\left(\frac{\sqrt{174}}{6}-\frac{3}{2}\right)\geq 0 and x-\left(-\frac{\sqrt{174}}{6}-\frac{3}{2}\right)\leq 0.

$$x-\left(\frac{\sqrt{174}}{6}-\frac{3}{2}\right)\geq 0 x-\left(-\frac{\sqrt{174}}{6}-\frac{3}{2}\right)\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(\frac{\sqrt{174}}{6}-\frac{3}{2}\right)\leq 0 and x-\left(-\frac{\sqrt{174}}{6}-\frac{3}{2}\right)\geq 0.

$$x-\left(-\frac{\sqrt{174}}{6}-\frac{3}{2}\right)\geq 0 x-\left(\frac{\sqrt{174}}{6}-\frac{3}{2}\right)\leq 0$$

The solution satisfying both inequalities is x\in \left[-\frac{\sqrt{174}}{6}-\frac{3}{2},\frac{\sqrt{174}}{6}-\frac{3}{2}\right].

$$x\in \begin{bmatrix}-\frac{\sqrt{174}}{6}-\frac{3}{2},\frac{\sqrt{174}}{6}-\frac{3}{2}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{\sqrt{174}}{6}-\frac{3}{2},\frac{\sqrt{174}}{6}-\frac{3}{2}\end{bmatrix}$$