Solve 9t^2+4+2tgeq16t

$$9 t ^ { 2 } + 4 + 2 t \geq 16 t$$

Solve for t

$$t\in (-\infty,\frac{7-\sqrt{13}}{9}]\cup [\frac{\sqrt{13}+7}{9},\infty)$$

Steps Using the Quadratic Formula

Subtract 16t from both sides.

$$9t^{2}+4+2t-16t\geq 0$$

Combine 2t and -16t to get -14t.

$$9t^{2}+4-14t\geq 0$$

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.

$$9t^{2}+4-14t=0$$

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 9 for a, -14 for b, and 4 for c in the quadratic formula.

$$t=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 9\times 4}}{2\times 9}$$

Do the calculations.

$$t=\frac{14±2\sqrt{13}}{18}$$

Solve the equation t=\frac{14±2\sqrt{13}}{18} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{13}+7}{9} t=\frac{7-\sqrt{13}}{9}$$

Rewrite the inequality by using the obtained solutions.

$$9\left(t-\frac{\sqrt{13}+7}{9}\right)\left(t-\frac{7-\sqrt{13}}{9}\right)\geq 0$$

For the product to be ≥0, t-\frac{\sqrt{13}+7}{9} and t-\frac{7-\sqrt{13}}{9} have to be both ≤0 or both ≥0. Consider the case when t-\frac{\sqrt{13}+7}{9} and t-\frac{7-\sqrt{13}}{9} are both ≤0.

$$t-\frac{\sqrt{13}+7}{9}\leq 0 t-\frac{7-\sqrt{13}}{9}\leq 0$$

The solution satisfying both inequalities is t\leq \frac{7-\sqrt{13}}{9}.

$$t\leq \frac{7-\sqrt{13}}{9}$$

Consider the case when t-\frac{\sqrt{13}+7}{9} and t-\frac{7-\sqrt{13}}{9} are both ≥0.

$$t-\frac{7-\sqrt{13}}{9}\geq 0 t-\frac{\sqrt{13}+7}{9}\geq 0$$

The solution satisfying both inequalities is t\geq \frac{\sqrt{13}+7}{9}.

$$t\geq \frac{\sqrt{13}+7}{9}$$

The final solution is the union of the obtained solutions.

$$t\leq \frac{7-\sqrt{13}}{9}\text{; }t\geq \frac{\sqrt{13}+7}{9}$$