Solve 3t+3-3t^2-2

$$3 t + 3 - 3 t ^ { 2 } - 2$$

Factor

$$-3\left(t-\left(-\frac{\sqrt{21}}{6}+\frac{1}{2}\right)\right)\left(t-\left(\frac{\sqrt{21}}{6}+\frac{1}{2}\right)\right)$$

Steps Using the Quadratic Formula

Subtract 2 from 3 to get 1.

$$factor(3t+1-3t^{2})$$

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.

$$-3t^{2}+3t+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

Square 3.

$$t=\frac{-3±\sqrt{9-4\left(-3\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$t=\frac{-3±\sqrt{9+12}}{2\left(-3\right)}$$

Add 9 to 12.

$$t=\frac{-3±\sqrt{21}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$t=\frac{-3±\sqrt{21}}{-6}$$

Now solve the equation t=\frac{-3±\sqrt{21}}{-6} when ± is plus. Add -3 to \sqrt{21}.

$$t=\frac{\sqrt{21}-3}{-6}$$

Divide -3+\sqrt{21} by -6.

$$t=-\frac{\sqrt{21}}{6}+\frac{1}{2}$$

Now solve the equation t=\frac{-3±\sqrt{21}}{-6} when ± is minus. Subtract \sqrt{21} from -3.

$$t=\frac{-\sqrt{21}-3}{-6}$$

Divide -3-\sqrt{21} by -6.

$$t=\frac{\sqrt{21}}{6}+\frac{1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{2}-\frac{\sqrt{21}}{6} for x_{1} and \frac{1}{2}+\frac{\sqrt{21}}{6} for x_{2}.

$$-3t^{2}+3t+1=-3\left(t-\left(-\frac{\sqrt{21}}{6}+\frac{1}{2}\right)\right)\left(t-\left(\frac{\sqrt{21}}{6}+\frac{1}{2}\right)\right)$$

Evaluate

$$1+3t-3t^{2}$$

Solution Steps

Subtract 2 from 3 to get 1.

$$3t+1-3t^{2}$$