Solve 36m^2-4(m+2)(4m-1)geq0

$$36 m ^ { 2 } - 4 ( m + 2 ) ( 4 m - 1 ) \geq 0$$

Solve for m

$$m\in (-\infty,\frac{2}{5}]\cup [1,\infty)$$

Steps Using the Quadratic Formula

Multiply -1 and 4 to get -4.

$$36m^{2}-4\left(m+2\right)\left(4m-1\right)\geq 0$$

Use the distributive property to multiply -4 by m+2.

$$36m^{2}+\left(-4m-8\right)\left(4m-1\right)\geq 0$$

Use the distributive property to multiply -4m-8 by 4m-1 and combine like terms.

$$36m^{2}-16m^{2}-28m+8\geq 0$$

Combine 36m^{2} and -16m^{2} to get 20m^{2}.

$$20m^{2}-28m+8\geq 0$$

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.

$$20m^{2}-28m+8=0$$

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 20 for a, -28 for b, and 8 for c in the quadratic formula.

$$m=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 20\times 8}}{2\times 20}$$

Do the calculations.

$$m=\frac{28±12}{40}$$

Solve the equation m=\frac{28±12}{40} when ± is plus and when ± is minus.

$$m=1 m=\frac{2}{5}$$

Rewrite the inequality by using the obtained solutions.

$$20\left(m-1\right)\left(m-\frac{2}{5}\right)\geq 0$$

For the product to be ≥0, m-1 and m-\frac{2}{5} have to be both ≤0 or both ≥0. Consider the case when m-1 and m-\frac{2}{5} are both ≤0.

$$m-1\leq 0 m-\frac{2}{5}\leq 0$$

The solution satisfying both inequalities is m\leq \frac{2}{5}.

$$m\leq \frac{2}{5}$$

Consider the case when m-1 and m-\frac{2}{5} are both ≥0.

$$m-\frac{2}{5}\geq 0 m-1\geq 0$$

The solution satisfying both inequalities is m\geq 1.

$$m\geq 1$$

The final solution is the union of the obtained solutions.

$$m\leq \frac{2}{5}\text{; }m\geq 1$$