Solve t^2-12t+52geq0

$$t ^ { 2 } - 12 t + 52 \geq 0$$

Solve for t

$$t\in \mathrm{R}$$

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.

$$t^{2}-12t+52=0$$

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, -12 for b, and 52 for c in the quadratic formula.

$$t=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 1\times 52}}{2}$$

Do the calculations.

$$t=\frac{12±\sqrt{-64}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression t^{2}-12t+52 has the same sign for any t. To determine the sign, calculate the value of the expression for t=0.

$$0^{2}-12\times 0+52=52$$

The value of the expression t^{2}-12t+52 is always positive. Inequality holds for t\in \mathrm{R}.

$$t\in \mathrm{R}$$