Solve 2=2t-3/2t^2

$$2 = 2 t - \frac { 3 } { 2 } t ^ { 2 }$$

Solve for t

$$t=\frac{2+2\sqrt{2}i}{3}\approx 0.666666667+0.942809042i$$

$$t=\frac{-2\sqrt{2}i+2}{3}\approx 0.666666667-0.942809042i$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$2t-\frac{3}{2}t^{2}=2$$

Subtract 2 from both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{3}{2} for a, 2 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 2.

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

Multiply -4 times -\frac{3}{2}.

$$t=\frac{-2±\sqrt{4+6\left(-2\right)}}{2\left(-\frac{3}{2}\right)}$$

Multiply 6 times -2.

$$t=\frac{-2±\sqrt{4-12}}{2\left(-\frac{3}{2}\right)}$$

Add 4 to -12.

$$t=\frac{-2±\sqrt{-8}}{2\left(-\frac{3}{2}\right)}$$

Take the square root of -8.

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

Multiply 2 times -\frac{3}{2}.

$$t=\frac{-2±2\sqrt{2}i}{-3}$$

Now solve the equation t=\frac{-2±2\sqrt{2}i}{-3} when ± is plus. Add -2 to 2i\sqrt{2}.

$$t=\frac{-2+2\sqrt{2}i}{-3}$$

Divide -2+2i\sqrt{2} by -3.

$$t=\frac{-2\sqrt{2}i+2}{3}$$

Now solve the equation t=\frac{-2±2\sqrt{2}i}{-3} when ± is minus. Subtract 2i\sqrt{2} from -2.

$$t=\frac{-2\sqrt{2}i-2}{-3}$$

Divide -2-2i\sqrt{2} by -3.

$$t=\frac{2+2\sqrt{2}i}{3}$$

The equation is now solved.

$$t=\frac{-2\sqrt{2}i+2}{3} t=\frac{2+2\sqrt{2}i}{3}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$2t-\frac{3}{2}t^{2}=2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides of the equation by -\frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

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

Dividing by -\frac{3}{2} undoes the multiplication by -\frac{3}{2}.

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

Divide 2 by -\frac{3}{2} by multiplying 2 by the reciprocal of -\frac{3}{2}.

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

Divide 2 by -\frac{3}{2} by multiplying 2 by the reciprocal of -\frac{3}{2}.

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

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{4}{3}t+\frac{4}{9}=-\frac{4}{3}+\frac{4}{9}$$

Add -\frac{4}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{4}{3}t+\frac{4}{9}=-\frac{8}{9}$$

Factor t^{2}-\frac{4}{3}t+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{2}{3}\right)^{2}=-\frac{8}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{2}{3}\right)^{2}}=\sqrt{-\frac{8}{9}}$$

Simplify.

$$t-\frac{2}{3}=\frac{2\sqrt{2}i}{3} t-\frac{2}{3}=-\frac{2\sqrt{2}i}{3}$$

Add \frac{2}{3} to both sides of the equation.

$$t=\frac{2+2\sqrt{2}i}{3} t=\frac{-2\sqrt{2}i+2}{3}$$