Solve 4x^2-4*1/3*(x+6)>0

$$4 { x }^{ 2 } -4 \times \frac{ 1 }{ 3 } \times (x+6) > 0$$

Solve for x

$$x\in \left(-\infty,\frac{1-\sqrt{73}}{6}\right)\cup \left(\frac{\sqrt{73}+1}{6},\infty\right)$$

Steps Using the Quadratic Formula

Multiply 4 and \frac{1}{3} to get \frac{4}{3}.

$$4x^{2}-\frac{4}{3}\left(x+6\right)>0$$

Use the distributive property to multiply -\frac{4}{3} by x+6.

$$4x^{2}-\frac{4}{3}x-8>0$$

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.

$$4x^{2}-\frac{4}{3}x-8=0$$

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 4 for a, -\frac{4}{3} for b, and -8 for c in the quadratic formula.

$$x=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\left(-\frac{4}{3}\right)^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{\frac{4}{3}±\frac{4}{3}\sqrt{73}}{8}$$

Solve the equation x=\frac{\frac{4}{3}±\frac{4}{3}\sqrt{73}}{8} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{73}+1}{6} x=\frac{1-\sqrt{73}}{6}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(x-\frac{\sqrt{73}+1}{6}\right)\left(x-\frac{1-\sqrt{73}}{6}\right)>0$$

For the product to be positive, x-\frac{\sqrt{73}+1}{6} and x-\frac{1-\sqrt{73}}{6} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{73}+1}{6} and x-\frac{1-\sqrt{73}}{6} are both negative.

$$x-\frac{\sqrt{73}+1}{6}<0 x-\frac{1-\sqrt{73}}{6}<0$$

The solution satisfying both inequalities is x<\frac{1-\sqrt{73}}{6}.

$$x<\frac{1-\sqrt{73}}{6}$$

Consider the case when x-\frac{\sqrt{73}+1}{6} and x-\frac{1-\sqrt{73}}{6} are both positive.

$$x-\frac{1-\sqrt{73}}{6}>0 x-\frac{\sqrt{73}+1}{6}>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{73}+1}{6}.

$$x>\frac{\sqrt{73}+1}{6}$$

The final solution is the union of the obtained solutions.

$$x<\frac{1-\sqrt{73}}{6}\text{; }x>\frac{\sqrt{73}+1}{6}$$