Solve -t^2-20+t<0

$$- t ^ { 2 } - 20 + t < 0$$

Solve for t

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

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -t^{2}-20+t positive. Since -1 is negative, the inequality direction is changed.

$$t^{2}+20-t>0$$

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.

$$t^{2}+20-t=0$$

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, -1 for b, and 20 for c in the quadratic formula.

$$t=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 20}}{2}$$

Do the calculations.

$$t=\frac{1±\sqrt{-79}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression t^{2}+20-t has the same sign for any t. To determine the sign, calculate the value of the expression for t=0.

$$0^{2}+20-0=20$$

The value of the expression t^{2}+20-t is always positive. Inequality holds for t\in \mathrm{R}.

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