Solve -x^2+7x+144<0

$$- x ^ { 2 } + 7 x + 144 < 0$$

Solve for x

$$x\in \left(-\infty,-9\right)\cup \left(16,\infty\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -x^{2}+7x+144 positive. Since -1 is negative, the inequality direction is changed.

$$x^{2}-7x-144>0$$

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.

$$x^{2}-7x-144=0$$

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, -7 for b, and -144 for c in the quadratic formula.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 1\left(-144\right)}}{2}$$

Do the calculations.

$$x=\frac{7±25}{2}$$

Solve the equation x=\frac{7±25}{2} when ± is plus and when ± is minus.

$$x=16 x=-9$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-16\right)\left(x+9\right)>0$$

For the product to be positive, x-16 and x+9 have to be both negative or both positive. Consider the case when x-16 and x+9 are both negative.

$$x-16<0 x+9<0$$

The solution satisfying both inequalities is x<-9.

$$x<-9$$

Consider the case when x-16 and x+9 are both positive.

$$x+9>0 x-16>0$$

The solution satisfying both inequalities is x>16.

$$x>16$$

The final solution is the union of the obtained solutions.

$$x<-9\text{; }x>16$$