Solve -2x^2+30x-72geq0

$$- 2 x ^ { 2 } + 30 x - 72 \geq 0$$

Solve for x

$$x\in \begin{bmatrix}3,12\end{bmatrix}$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -2x^{2}+30x-72 positive. Since -1 is negative, the inequality direction is changed.

$$2x^{2}-30x+72\leq 0$$

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.

$$2x^{2}-30x+72=0$$

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, -30 for b, and 72 for c in the quadratic formula.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 2\times 72}}{2\times 2}$$

Do the calculations.

$$x=\frac{30±18}{4}$$

Solve the equation x=\frac{30±18}{4} when ± is plus and when ± is minus.

$$x=12 x=3$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-12\right)\left(x-3\right)\leq 0$$

For the product to be ≤0, one of the values x-12 and x-3 has to be ≥0 and the other has to be ≤0. Consider the case when x-12\geq 0 and x-3\leq 0.

$$x-12\geq 0 x-3\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-12\leq 0 and x-3\geq 0.

$$x-3\geq 0 x-12\leq 0$$

The solution satisfying both inequalities is x\in \left[3,12\right].

$$x\in \begin{bmatrix}3,12\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}3,12\end{bmatrix}$$