Solve -2p^2+12p-12<0

$$- 2 p ^ { 2 } + 12 p - 12 < 0$$

Solve for p

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

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -2p^{2}+12p-12 positive. Since -1 is negative, the inequality direction is changed.

$$2p^{2}-12p+12>0$$

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.

$$2p^{2}-12p+12=0$$

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

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

Do the calculations.

$$p=\frac{12±4\sqrt{3}}{4}$$

Solve the equation p=\frac{12±4\sqrt{3}}{4} when ± is plus and when ± is minus.

$$p=\sqrt{3}+3 p=3-\sqrt{3}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(p-\left(\sqrt{3}+3\right)\right)\left(p-\left(3-\sqrt{3}\right)\right)>0$$

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

$$p-\left(\sqrt{3}+3\right)<0 p-\left(3-\sqrt{3}\right)<0$$

The solution satisfying both inequalities is p<3-\sqrt{3}.

$$p<3-\sqrt{3}$$

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

$$p-\left(3-\sqrt{3}\right)>0 p-\left(\sqrt{3}+3\right)>0$$

The solution satisfying both inequalities is p>\sqrt{3}+3.

$$p>\sqrt{3}+3$$

The final solution is the union of the obtained solutions.

$$p<3-\sqrt{3}\text{; }p>\sqrt{3}+3$$