Solve (6sqrt{3}m+10sqrt{3})^2-4*12(3m^2+10m+3)geq0

$$( 6 \sqrt { 3 } m + 10 \sqrt { 3 } ) ^ { 2 } - 4 \cdot 12 ( 3 m ^ { 2 } + 10 m + 3 ) \geq 0$$

Solve for m

$$m\in \begin{bmatrix}-\frac{13}{3},1\end{bmatrix}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6\sqrt{3}m+10\sqrt{3}\right)^{2}.

$$36\left(\sqrt{3}\right)^{2}m^{2}+120\sqrt{3}m\sqrt{3}+100\left(\sqrt{3}\right)^{2}-4\times 12\left(3m^{2}+10m+3\right)\geq 0$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$36\left(\sqrt{3}\right)^{2}m^{2}+120\times 3m+100\left(\sqrt{3}\right)^{2}-4\times 12\left(3m^{2}+10m+3\right)\geq 0$$

The square of \sqrt{3} is 3.

$$36\times 3m^{2}+120\times 3m+100\left(\sqrt{3}\right)^{2}-4\times 12\left(3m^{2}+10m+3\right)\geq 0$$

Multiply 36 and 3 to get 108.

$$108m^{2}+120\times 3m+100\left(\sqrt{3}\right)^{2}-4\times 12\left(3m^{2}+10m+3\right)\geq 0$$

Multiply 120 and 3 to get 360.

$$108m^{2}+360m+100\left(\sqrt{3}\right)^{2}-4\times 12\left(3m^{2}+10m+3\right)\geq 0$$

The square of \sqrt{3} is 3.

$$108m^{2}+360m+100\times 3-4\times 12\left(3m^{2}+10m+3\right)\geq 0$$

Multiply 100 and 3 to get 300.

$$108m^{2}+360m+300-4\times 12\left(3m^{2}+10m+3\right)\geq 0$$

Multiply 4 and 12 to get 48.

$$108m^{2}+360m+300-48\left(3m^{2}+10m+3\right)\geq 0$$

Use the distributive property to multiply -48 by 3m^{2}+10m+3.

$$108m^{2}+360m+300-144m^{2}-480m-144\geq 0$$

Combine 108m^{2} and -144m^{2} to get -36m^{2}.

$$-36m^{2}+360m+300-480m-144\geq 0$$

Combine 360m and -480m to get -120m.

$$-36m^{2}-120m+300-144\geq 0$$

Subtract 144 from 300 to get 156.

$$-36m^{2}-120m+156\geq 0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -36m^{2}-120m+156 positive. Since -1 is negative, the inequality direction is changed.

$$36m^{2}+120m-156\leq 0$$

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.

$$36m^{2}+120m-156=0$$

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 36 for a, 120 for b, and -156 for c in the quadratic formula.

$$m=\frac{-120±\sqrt{120^{2}-4\times 36\left(-156\right)}}{2\times 36}$$

Do the calculations.

$$m=\frac{-120±192}{72}$$

Solve the equation m=\frac{-120±192}{72} when ± is plus and when ± is minus.

$$m=1 m=-\frac{13}{3}$$

Rewrite the inequality by using the obtained solutions.

$$36\left(m-1\right)\left(m+\frac{13}{3}\right)\leq 0$$

For the product to be ≤0, one of the values m-1 and m+\frac{13}{3} has to be ≥0 and the other has to be ≤0. Consider the case when m-1\geq 0 and m+\frac{13}{3}\leq 0.

$$m-1\geq 0 m+\frac{13}{3}\leq 0$$

This is false for any m.

$$m\in \emptyset$$

Consider the case when m-1\leq 0 and m+\frac{13}{3}\geq 0.

$$m+\frac{13}{3}\geq 0 m-1\leq 0$$

The solution satisfying both inequalities is m\in \left[-\frac{13}{3},1\right].

$$m\in \begin{bmatrix}-\frac{13}{3},1\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$m\in \begin{bmatrix}-\frac{13}{3},1\end{bmatrix}$$