Solve 8-3x-x^2>0

$$8 - 3 x - x ^ { 2 } > 0$$

Solve for x

$$x\in \left(\frac{-\sqrt{41}-3}{2},\frac{\sqrt{41}-3}{2}\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in 8-3x-x^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-8+3x+x^{2}<0$$

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.

$$-8+3x+x^{2}=0$$

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, 3 for b, and -8 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\left(-8\right)}}{2}$$

Do the calculations.

$$x=\frac{-3±\sqrt{41}}{2}$$

Solve the equation x=\frac{-3±\sqrt{41}}{2} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{41}-3}{2} x=\frac{-\sqrt{41}-3}{2}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\frac{\sqrt{41}-3}{2}\right)\left(x-\frac{-\sqrt{41}-3}{2}\right)<0$$

For the product to be negative, x-\frac{\sqrt{41}-3}{2} and x-\frac{-\sqrt{41}-3}{2} have to be of the opposite signs. Consider the case when x-\frac{\sqrt{41}-3}{2} is positive and x-\frac{-\sqrt{41}-3}{2} is negative.

$$x-\frac{\sqrt{41}-3}{2}>0 x-\frac{-\sqrt{41}-3}{2}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{-\sqrt{41}-3}{2} is positive and x-\frac{\sqrt{41}-3}{2} is negative.

$$x-\frac{-\sqrt{41}-3}{2}>0 x-\frac{\sqrt{41}-3}{2}<0$$

The solution satisfying both inequalities is x\in \left(\frac{-\sqrt{41}-3}{2},\frac{\sqrt{41}-3}{2}\right).

$$x\in \left(\frac{-\sqrt{41}-3}{2},\frac{\sqrt{41}-3}{2}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{-\sqrt{41}-3}{2},\frac{\sqrt{41}-3}{2}\right)$$