Solve 36a^2-4*9(a+2)>0

$$36 a ^ { 2 } - 4 \times 9 ( a + 2 ) > 0$$

Solve for a

$$a\in \left(-\infty,-1\right)\cup \left(2,\infty\right)$$

Steps Using the Quadratic Formula

Multiply 4 and 9 to get 36.

$$36a^{2}-36\left(a+2\right)>0$$

Use the distributive property to multiply -36 by a+2.

$$36a^{2}-36a-72>0$$

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.

$$36a^{2}-36a-72=0$$

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 36 for a, -36 for b, and -72 for c in the quadratic formula.

$$a=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 36\left(-72\right)}}{2\times 36}$$

Do the calculations.

$$a=\frac{36±108}{72}$$

Solve the equation a=\frac{36±108}{72} when ± is plus and when ± is minus.

$$a=2 a=-1$$

Rewrite the inequality by using the obtained solutions.

$$36\left(a-2\right)\left(a+1\right)>0$$

For the product to be positive, a-2 and a+1 have to be both negative or both positive. Consider the case when a-2 and a+1 are both negative.

$$a-2<0 a+1<0$$

The solution satisfying both inequalities is a<-1.

$$a<-1$$

Consider the case when a-2 and a+1 are both positive.

$$a+1>0 a-2>0$$

The solution satisfying both inequalities is a>2.

$$a>2$$

The final solution is the union of the obtained solutions.

$$a<-1\text{; }a>2$$