Solve (2-a)^2-16>0

$$( 2 - a ) ^ { 2 } - 16 > 0$$

Solve for a

$$a\in \left(-\infty,-2\right)\cup \left(6,\infty\right)$$

Steps Using the Quadratic Formula

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

$$4-4a+a^{2}-16>0$$

Subtract 16 from 4 to get -12.

$$-12-4a+a^{2}>0$$

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.

$$-12-4a+a^{2}=0$$

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 1 for a, -4 for b, and -12 for c in the quadratic formula.

$$a=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-12\right)}}{2}$$

Do the calculations.

$$a=\frac{4±8}{2}$$

Solve the equation a=\frac{4±8}{2} when ± is plus and when ± is minus.

$$a=6 a=-2$$

Rewrite the inequality by using the obtained solutions.

$$\left(a-6\right)\left(a+2\right)>0$$

For the product to be positive, a-6 and a+2 have to be both negative or both positive. Consider the case when a-6 and a+2 are both negative.

$$a-6<0 a+2<0$$

The solution satisfying both inequalities is a<-2.

$$a<-2$$

Consider the case when a-6 and a+2 are both positive.

$$a+2>0 a-6>0$$

The solution satisfying both inequalities is a>6.

$$a>6$$

The final solution is the union of the obtained solutions.

$$a<-2\text{; }a>6$$