Solve x^2+3x-40>0

$$x ^ { 2 } + 3 x - 40 > 0$$

Solve for x

$$x\in \left(-\infty,-8\right)\cup \left(5,\infty\right)$$

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.

$$x^{2}+3x-40=0$$

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 -40 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\left(-40\right)}}{2}$$

Do the calculations.

$$x=\frac{-3±13}{2}$$

Solve the equation x=\frac{-3±13}{2} when ± is plus and when ± is minus.

$$x=5 x=-8$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-5\right)\left(x+8\right)>0$$

For the product to be positive, x-5 and x+8 have to be both negative or both positive. Consider the case when x-5 and x+8 are both negative.

$$x-5<0 x+8<0$$

The solution satisfying both inequalities is x<-8.

$$x<-8$$

Consider the case when x-5 and x+8 are both positive.

$$x+8>0 x-5>0$$

The solution satisfying both inequalities is x>5.

$$x>5$$

The final solution is the union of the obtained solutions.

$$x<-8\text{; }x>5$$