Solve 3+6x-x^2>0

$$3+6x- { x }^{ 2 } > 0$$

Solve for x

$$x\in \left(3-2\sqrt{3},2\sqrt{3}+3\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in 3+6x-x^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-3-6x+x^{2}<0$$

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.

$$-3-6x+x^{2}=0$$

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, -6 for b, and -3 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\left(-3\right)}}{2}$$

Do the calculations.

$$x=\frac{6±4\sqrt{3}}{2}$$

Solve the equation x=\frac{6±4\sqrt{3}}{2} when ± is plus and when ± is minus.

$$x=2\sqrt{3}+3 x=3-2\sqrt{3}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(2\sqrt{3}+3\right)\right)\left(x-\left(3-2\sqrt{3}\right)\right)<0$$

For the product to be negative, x-\left(2\sqrt{3}+3\right) and x-\left(3-2\sqrt{3}\right) have to be of the opposite signs. Consider the case when x-\left(2\sqrt{3}+3\right) is positive and x-\left(3-2\sqrt{3}\right) is negative.

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

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(3-2\sqrt{3}\right) is positive and x-\left(2\sqrt{3}+3\right) is negative.

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

The solution satisfying both inequalities is x\in \left(3-2\sqrt{3},2\sqrt{3}+3\right).

$$x\in \left(3-2\sqrt{3},2\sqrt{3}+3\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(3-2\sqrt{3},2\sqrt{3}+3\right)$$