Solve -x^2+8x+12>0

$$- { x }^{ 2 } +8x+12 > 0$$

Solve for x

$$x\in \left(4-2\sqrt{7},2\sqrt{7}+4\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -x^{2}+8x+12 positive. Since -1 is negative, the inequality direction is changed.

$$x^{2}-8x-12<0$$

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.

$$x^{2}-8x-12=0$$

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 1 for a, -8 for b, and -12 for c in the quadratic formula.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 1\left(-12\right)}}{2}$$

Do the calculations.

$$x=\frac{8±4\sqrt{7}}{2}$$

Solve the equation x=\frac{8±4\sqrt{7}}{2} when ± is plus and when ± is minus.

$$x=2\sqrt{7}+4 x=4-2\sqrt{7}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(2\sqrt{7}+4\right)\right)\left(x-\left(4-2\sqrt{7}\right)\right)<0$$

For the product to be negative, x-\left(2\sqrt{7}+4\right) and x-\left(4-2\sqrt{7}\right) have to be of the opposite signs. Consider the case when x-\left(2\sqrt{7}+4\right) is positive and x-\left(4-2\sqrt{7}\right) is negative.

$$x-\left(2\sqrt{7}+4\right)>0 x-\left(4-2\sqrt{7}\right)<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(4-2\sqrt{7}\right) is positive and x-\left(2\sqrt{7}+4\right) is negative.

$$x-\left(4-2\sqrt{7}\right)>0 x-\left(2\sqrt{7}+4\right)<0$$

The solution satisfying both inequalities is x\in \left(4-2\sqrt{7},2\sqrt{7}+4\right).

$$x\in \left(4-2\sqrt{7},2\sqrt{7}+4\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(4-2\sqrt{7},2\sqrt{7}+4\right)$$