Solve t^2-2t+3/4geq0

$$t ^ { 2 } - 2 t + \frac { 3 } { 4 } \geq 0$$

Solve for t

$$t\in (-\infty,\frac{1}{2}]\cup [\frac{3}{2},\infty)$$

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.

$$t^{2}-2t+\frac{3}{4}=0$$

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, -2 for b, and \frac{3}{4} for c in the quadratic formula.

$$t=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\times \frac{3}{4}}}{2}$$

Do the calculations.

$$t=\frac{2±1}{2}$$

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

$$t=\frac{3}{2} t=\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$\left(t-\frac{3}{2}\right)\left(t-\frac{1}{2}\right)\geq 0$$

For the product to be ≥0, t-\frac{3}{2} and t-\frac{1}{2} have to be both ≤0 or both ≥0. Consider the case when t-\frac{3}{2} and t-\frac{1}{2} are both ≤0.

$$t-\frac{3}{2}\leq 0 t-\frac{1}{2}\leq 0$$

The solution satisfying both inequalities is t\leq \frac{1}{2}.

$$t\leq \frac{1}{2}$$

Consider the case when t-\frac{3}{2} and t-\frac{1}{2} are both ≥0.

$$t-\frac{1}{2}\geq 0 t-\frac{3}{2}\geq 0$$

The solution satisfying both inequalities is t\geq \frac{3}{2}.

$$t\geq \frac{3}{2}$$

The final solution is the union of the obtained solutions.

$$t\leq \frac{1}{2}\text{; }t\geq \frac{3}{2}$$