Solve 1/2x^2+x-12>0

$$\frac { 1 } { 2 } x ^ { 2 } + x - 12 > 0$$

Solve for x

$$x\in \left(-\infty,-6\right)\cup \left(4,\infty\right)$$

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.

$$\frac{1}{2}x^{2}+x-12=0$$

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 \frac{1}{2} for a, 1 for b, and -12 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-1±5}{1}$$

Solve the equation x=\frac{-1±5}{1} when ± is plus and when ± is minus.

$$x=4 x=-6$$

Rewrite the inequality by using the obtained solutions.

$$\frac{1}{2}\left(x-4\right)\left(x+6\right)>0$$

For the product to be positive, x-4 and x+6 have to be both negative or both positive. Consider the case when x-4 and x+6 are both negative.

$$x-4<0 x+6<0$$

The solution satisfying both inequalities is x<-6.

$$x<-6$$

Consider the case when x-4 and x+6 are both positive.

$$x+6>0 x-4>0$$

The solution satisfying both inequalities is x>4.

$$x>4$$

The final solution is the union of the obtained solutions.

$$x<-6\text{; }x>4$$