Solve x^2+x-12>0

$$x ^ { 2 } + x - 12 > 0$$

Solve for x

$$x\in \left(-\infty,-4\right)\cup \left(3,\infty\right)$$

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.

$$x^{2}+x-12=0$$

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, 1 for b, and -12 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\left(-12\right)}}{2}$$

Do the calculations.

$$x=\frac{-1±7}{2}$$

Solve the equation x=\frac{-1±7}{2} when ± is plus and when ± is minus.

$$x=3 x=-4$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-3\right)\left(x+4\right)>0$$

For the product to be positive, x-3 and x+4 have to be both negative or both positive. Consider the case when x-3 and x+4 are both negative.

$$x-3<0 x+4<0$$

The solution satisfying both inequalities is x<-4.

$$x<-4$$

Consider the case when x-3 and x+4 are both positive.

$$x+4>0 x-3>0$$

The solution satisfying both inequalities is x>3.

$$x>3$$

The final solution is the union of the obtained solutions.

$$x<-4\text{; }x>3$$