Solve 8x-15>x^2

$$8 x - 15 > x ^ { 2 }$$

Solve for x

$$x\in \left(3,5\right)$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$8x-15-x^{2}>0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 8x-15-x^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-8x+15+x^{2}<0$$

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.

$$-8x+15+x^{2}=0$$

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, -8 for b, and 15 for c in the quadratic formula.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 1\times 15}}{2}$$

Do the calculations.

$$x=\frac{8±2}{2}$$

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

$$x=5 x=3$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-5\right)\left(x-3\right)<0$$

For the product to be negative, x-5 and x-3 have to be of the opposite signs. Consider the case when x-5 is positive and x-3 is negative.

$$x-5>0 x-3<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-3 is positive and x-5 is negative.

$$x-3>0 x-5<0$$

The solution satisfying both inequalities is x\in \left(3,5\right).

$$x\in \left(3,5\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(3,5\right)$$