Solve a^2+3a-60>0

$$a ^ { 2 } + 3 a - 60 > 0$$

Solve for a

$$a\in \left(-\infty,\frac{-\sqrt{249}-3}{2}\right)\cup \left(\frac{\sqrt{249}-3}{2},\infty\right)$$

Steps Using the Quadratic Formula

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.

$$a^{2}+3a-60=0$$

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 -60 for c in the quadratic formula.

$$a=\frac{-3±\sqrt{3^{2}-4\times 1\left(-60\right)}}{2}$$

Do the calculations.

$$a=\frac{-3±\sqrt{249}}{2}$$

Solve the equation a=\frac{-3±\sqrt{249}}{2} when ± is plus and when ± is minus.

$$a=\frac{\sqrt{249}-3}{2} a=\frac{-\sqrt{249}-3}{2}$$

Rewrite the inequality by using the obtained solutions.

$$\left(a-\frac{\sqrt{249}-3}{2}\right)\left(a-\frac{-\sqrt{249}-3}{2}\right)>0$$

For the product to be positive, a-\frac{\sqrt{249}-3}{2} and a-\frac{-\sqrt{249}-3}{2} have to be both negative or both positive. Consider the case when a-\frac{\sqrt{249}-3}{2} and a-\frac{-\sqrt{249}-3}{2} are both negative.

$$a-\frac{\sqrt{249}-3}{2}<0 a-\frac{-\sqrt{249}-3}{2}<0$$

The solution satisfying both inequalities is a<\frac{-\sqrt{249}-3}{2}.

$$a<\frac{-\sqrt{249}-3}{2}$$

Consider the case when a-\frac{\sqrt{249}-3}{2} and a-\frac{-\sqrt{249}-3}{2} are both positive.

$$a-\frac{-\sqrt{249}-3}{2}>0 a-\frac{\sqrt{249}-3}{2}>0$$

The solution satisfying both inequalities is a>\frac{\sqrt{249}-3}{2}.

$$a>\frac{\sqrt{249}-3}{2}$$

The final solution is the union of the obtained solutions.

$$a<\frac{-\sqrt{249}-3}{2}\text{; }a>\frac{\sqrt{249}-3}{2}$$