Solve 14-2x^2-3xgeq0

$$14 - 2 x ^ { 2 } - 3 x \geq 0$$

Solve for x

$$x\in \begin{bmatrix}-\frac{7}{2},2\end{bmatrix}$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in 14-2x^{2}-3x positive. Since -1 is negative, the inequality direction is changed.

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

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.

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

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 2 for a, 3 for b, and -14 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-3±11}{4}$$

Solve the equation x=\frac{-3±11}{4} when ± is plus and when ± is minus.

$$x=2 x=-\frac{7}{2}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-2\right)\left(x+\frac{7}{2}\right)\leq 0$$

For the product to be ≤0, one of the values x-2 and x+\frac{7}{2} has to be ≥0 and the other has to be ≤0. Consider the case when x-2\geq 0 and x+\frac{7}{2}\leq 0.

$$x-2\geq 0 x+\frac{7}{2}\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-2\leq 0 and x+\frac{7}{2}\geq 0.

$$x+\frac{7}{2}\geq 0 x-2\leq 0$$

The solution satisfying both inequalities is x\in \left[-\frac{7}{2},2\right].

$$x\in \begin{bmatrix}-\frac{7}{2},2\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{7}{2},2\end{bmatrix}$$