Solve 4left(x+3right)^2-12x-72>0

$$4 { \left(x+3 \right) }^{ 2 } -12x-72 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{-3\sqrt{5}-3}{2}\right)\cup \left(\frac{3\sqrt{5}-3}{2},\infty\right)$$

Steps Using the Quadratic Formula

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

$$4\left(x^{2}+6x+9\right)-12x-72>0$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36-12x-72>0$$

Combine 24x and -12x to get 12x.

$$4x^{2}+12x+36-72>0$$

Subtract 72 from 36 to get -36.

$$4x^{2}+12x-36>0$$

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.

$$4x^{2}+12x-36=0$$

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, 12 for b, and -36 for c in the quadratic formula.

$$x=\frac{-12±\sqrt{12^{2}-4\times 4\left(-36\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{-12±12\sqrt{5}}{8}$$

Solve the equation x=\frac{-12±12\sqrt{5}}{8} when ± is plus and when ± is minus.

$$x=\frac{3\sqrt{5}-3}{2} x=\frac{-3\sqrt{5}-3}{2}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(x-\frac{3\sqrt{5}-3}{2}\right)\left(x-\frac{-3\sqrt{5}-3}{2}\right)>0$$

For the product to be positive, x-\frac{3\sqrt{5}-3}{2} and x-\frac{-3\sqrt{5}-3}{2} have to be both negative or both positive. Consider the case when x-\frac{3\sqrt{5}-3}{2} and x-\frac{-3\sqrt{5}-3}{2} are both negative.

$$x-\frac{3\sqrt{5}-3}{2}<0 x-\frac{-3\sqrt{5}-3}{2}<0$$

The solution satisfying both inequalities is x<\frac{-3\sqrt{5}-3}{2}.

$$x<\frac{-3\sqrt{5}-3}{2}$$

Consider the case when x-\frac{3\sqrt{5}-3}{2} and x-\frac{-3\sqrt{5}-3}{2} are both positive.

$$x-\frac{-3\sqrt{5}-3}{2}>0 x-\frac{3\sqrt{5}-3}{2}>0$$

The solution satisfying both inequalities is x>\frac{3\sqrt{5}-3}{2}.

$$x>\frac{3\sqrt{5}-3}{2}$$

The final solution is the union of the obtained solutions.

$$x<\frac{-3\sqrt{5}-3}{2}\text{; }x>\frac{3\sqrt{5}-3}{2}$$