Solve 5a^2-5a-2geq0

$$5 a ^ { 2 } - 5 a - 2 \geq 0$$

Solve for a

$$a\in (-\infty,-\frac{\sqrt{65}}{10}+\frac{1}{2}]\cup [\frac{\sqrt{65}}{10}+\frac{1}{2},\infty)$$

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.

$$5a^{2}-5a-2=0$$

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 5 for a, -5 for b, and -2 for c in the quadratic formula.

$$a=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 5\left(-2\right)}}{2\times 5}$$

Do the calculations.

$$a=\frac{5±\sqrt{65}}{10}$$

Solve the equation a=\frac{5±\sqrt{65}}{10} when ± is plus and when ± is minus.

$$a=\frac{\sqrt{65}}{10}+\frac{1}{2} a=-\frac{\sqrt{65}}{10}+\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$5\left(a-\left(\frac{\sqrt{65}}{10}+\frac{1}{2}\right)\right)\left(a-\left(-\frac{\sqrt{65}}{10}+\frac{1}{2}\right)\right)\geq 0$$

For the product to be ≥0, a-\left(\frac{\sqrt{65}}{10}+\frac{1}{2}\right) and a-\left(-\frac{\sqrt{65}}{10}+\frac{1}{2}\right) have to be both ≤0 or both ≥0. Consider the case when a-\left(\frac{\sqrt{65}}{10}+\frac{1}{2}\right) and a-\left(-\frac{\sqrt{65}}{10}+\frac{1}{2}\right) are both ≤0.

$$a-\left(\frac{\sqrt{65}}{10}+\frac{1}{2}\right)\leq 0 a-\left(-\frac{\sqrt{65}}{10}+\frac{1}{2}\right)\leq 0$$

The solution satisfying both inequalities is a\leq -\frac{\sqrt{65}}{10}+\frac{1}{2}.

$$a\leq -\frac{\sqrt{65}}{10}+\frac{1}{2}$$

Consider the case when a-\left(\frac{\sqrt{65}}{10}+\frac{1}{2}\right) and a-\left(-\frac{\sqrt{65}}{10}+\frac{1}{2}\right) are both ≥0.

$$a-\left(-\frac{\sqrt{65}}{10}+\frac{1}{2}\right)\geq 0 a-\left(\frac{\sqrt{65}}{10}+\frac{1}{2}\right)\geq 0$$

The solution satisfying both inequalities is a\geq \frac{\sqrt{65}}{10}+\frac{1}{2}.

$$a\geq \frac{\sqrt{65}}{10}+\frac{1}{2}$$

The final solution is the union of the obtained solutions.

$$a\leq -\frac{\sqrt{65}}{10}+\frac{1}{2}\text{; }a\geq \frac{\sqrt{65}}{10}+\frac{1}{2}$$