Solve 169a^2+456a+144geq0

$$169 a ^ { 2 } + 456 a + 144 \geq 0$$

Solve for a

$$a\in (-\infty,\frac{-96\sqrt{3}-228}{169}]\cup [\frac{96\sqrt{3}-228}{169},\infty)$$

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.

$$169a^{2}+456a+144=0$$

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 169 for a, 456 for b, and 144 for c in the quadratic formula.

$$a=\frac{-456±\sqrt{456^{2}-4\times 169\times 144}}{2\times 169}$$

Do the calculations.

$$a=\frac{-456±192\sqrt{3}}{338}$$

Solve the equation a=\frac{-456±192\sqrt{3}}{338} when ± is plus and when ± is minus.

$$a=\frac{96\sqrt{3}-228}{169} a=\frac{-96\sqrt{3}-228}{169}$$

Rewrite the inequality by using the obtained solutions.

$$169\left(a-\frac{96\sqrt{3}-228}{169}\right)\left(a-\frac{-96\sqrt{3}-228}{169}\right)\geq 0$$

For the product to be ≥0, a-\frac{96\sqrt{3}-228}{169} and a-\frac{-96\sqrt{3}-228}{169} have to be both ≤0 or both ≥0. Consider the case when a-\frac{96\sqrt{3}-228}{169} and a-\frac{-96\sqrt{3}-228}{169} are both ≤0.

$$a-\frac{96\sqrt{3}-228}{169}\leq 0 a-\frac{-96\sqrt{3}-228}{169}\leq 0$$

The solution satisfying both inequalities is a\leq \frac{-96\sqrt{3}-228}{169}.

$$a\leq \frac{-96\sqrt{3}-228}{169}$$

Consider the case when a-\frac{96\sqrt{3}-228}{169} and a-\frac{-96\sqrt{3}-228}{169} are both ≥0.

$$a-\frac{-96\sqrt{3}-228}{169}\geq 0 a-\frac{96\sqrt{3}-228}{169}\geq 0$$

The solution satisfying both inequalities is a\geq \frac{96\sqrt{3}-228}{169}.

$$a\geq \frac{96\sqrt{3}-228}{169}$$

The final solution is the union of the obtained solutions.

$$a\leq \frac{-96\sqrt{3}-228}{169}\text{; }a\geq \frac{96\sqrt{3}-228}{169}$$