Solve 4x+5<x^2

$$4 x + 5 < x ^ { 2 }$$

Solve for x

$$x\in \left(-\infty,-1\right)\cup \left(5,\infty\right)$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$4x+5-x^{2}<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 4x+5-x^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-4x-5+x^{2}>0$$

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.

$$-4x-5+x^{2}=0$$

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, -4 for b, and -5 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-5\right)}}{2}$$

Do the calculations.

$$x=\frac{4±6}{2}$$

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

$$x=5 x=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-5\right)\left(x+1\right)>0$$

For the product to be positive, x-5 and x+1 have to be both negative or both positive. Consider the case when x-5 and x+1 are both negative.

$$x-5<0 x+1<0$$

The solution satisfying both inequalities is x<-1.

$$x<-1$$

Consider the case when x-5 and x+1 are both positive.

$$x+1>0 x-5>0$$

The solution satisfying both inequalities is x>5.

$$x>5$$

The final solution is the union of the obtained solutions.

$$x<-1\text{; }x>5$$