Solve a^2+1/2a-1/2>0

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

Solve for a

$$a\in \left(-\infty,-1\right)\cup \left(\frac{1}{2},\infty\right)$$

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.

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

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

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

Do the calculations.

$$a=\frac{-\frac{1}{2}±\frac{3}{2}}{2}$$

Solve the equation a=\frac{-\frac{1}{2}±\frac{3}{2}}{2} when ± is plus and when ± is minus.

$$a=\frac{1}{2} a=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(a-\frac{1}{2}\right)\left(a+1\right)>0$$

For the product to be positive, a-\frac{1}{2} and a+1 have to be both negative or both positive. Consider the case when a-\frac{1}{2} and a+1 are both negative.

$$a-\frac{1}{2}<0 a+1<0$$

The solution satisfying both inequalities is a<-1.

$$a<-1$$

Consider the case when a-\frac{1}{2} and a+1 are both positive.

$$a+1>0 a-\frac{1}{2}>0$$

The solution satisfying both inequalities is a>\frac{1}{2}.

$$a>\frac{1}{2}$$

The final solution is the union of the obtained solutions.

$$a<-1\text{; }a>\frac{1}{2}$$