Solve (a-12)^2-8(16-a)>0

$$( a - 12 ) ^ { 2 } - 8 ( 16 - a ) > 0$$

Solve for a

$$a\in \left(-\infty,8-4\sqrt{3}\right)\cup \left(4\sqrt{3}+8,\infty\right)$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-12\right)^{2}.

$$a^{2}-24a+144-8\left(16-a\right)>0$$

Use the distributive property to multiply -8 by 16-a.

$$a^{2}-24a+144-128+8a>0$$

Subtract 128 from 144 to get 16.

$$a^{2}-24a+16+8a>0$$

Combine -24a and 8a to get -16a.

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

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.

$$a^{2}-16a+16=0$$

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, -16 for b, and 16 for c in the quadratic formula.

$$a=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$a=\frac{16±8\sqrt{3}}{2}$$

Solve the equation a=\frac{16±8\sqrt{3}}{2} when ± is plus and when ± is minus.

$$a=4\sqrt{3}+8 a=8-4\sqrt{3}$$

Rewrite the inequality by using the obtained solutions.

$$\left(a-\left(4\sqrt{3}+8\right)\right)\left(a-\left(8-4\sqrt{3}\right)\right)>0$$

For the product to be positive, a-\left(4\sqrt{3}+8\right) and a-\left(8-4\sqrt{3}\right) have to be both negative or both positive. Consider the case when a-\left(4\sqrt{3}+8\right) and a-\left(8-4\sqrt{3}\right) are both negative.

$$a-\left(4\sqrt{3}+8\right)<0 a-\left(8-4\sqrt{3}\right)<0$$

The solution satisfying both inequalities is a<8-4\sqrt{3}.

$$a<8-4\sqrt{3}$$

Consider the case when a-\left(4\sqrt{3}+8\right) and a-\left(8-4\sqrt{3}\right) are both positive.

$$a-\left(8-4\sqrt{3}\right)>0 a-\left(4\sqrt{3}+8\right)>0$$

The solution satisfying both inequalities is a>4\sqrt{3}+8.

$$a>4\sqrt{3}+8$$

The final solution is the union of the obtained solutions.

$$a<8-4\sqrt{3}\text{; }a>4\sqrt{3}+8$$