Solve 12+4a-a^2>0

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

Solve for a

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

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in 12+4a-a^{2} positive. Since -1 is negative, the inequality direction is changed.

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

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 negative, a-6 and a+2 have to be of the opposite signs. Consider the case when a-6 is positive and a+2 is negative.

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

This is false for any a.

$$a\in \emptyset$$

Consider the case when a+2 is positive and a-6 is negative.

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

The solution satisfying both inequalities is a\in \left(-2,6\right).

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

The final solution is the union of the obtained solutions.

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