Solve t^2-2t-3=-4+t-3

$$t ^ { 2 } - 2 t - 3 = - 4 + t - 3$$

Solve for t

$$t=\frac{3+\sqrt{7}i}{2}\approx 1.5+1.322875656i$$

$$t=\frac{-\sqrt{7}i+3}{2}\approx 1.5-1.322875656i$$

Steps Using the Quadratic Formula

Subtract 3 from -4 to get -7.

$$t^{2}-2t-3=-7+t$$

Subtract -7 from both sides.

$$t^{2}-2t-3-\left(-7\right)=t$$

The opposite of -7 is 7.

$$t^{2}-2t-3+7=t$$

Subtract t from both sides.

$$t^{2}-2t-3+7-t=0$$

Add -3 and 7 to get 4.

$$t^{2}-2t+4-t=0$$

Combine -2t and -t to get -3t.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -3.

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

Multiply -4 times 4.

$$t=\frac{-\left(-3\right)±\sqrt{9-16}}{2}$$

Add 9 to -16.

$$t=\frac{-\left(-3\right)±\sqrt{-7}}{2}$$

Take the square root of -7.

$$t=\frac{-\left(-3\right)±\sqrt{7}i}{2}$$

The opposite of -3 is 3.

$$t=\frac{3±\sqrt{7}i}{2}$$

Now solve the equation t=\frac{3±\sqrt{7}i}{2} when ± is plus. Add 3 to i\sqrt{7}.

$$t=\frac{3+\sqrt{7}i}{2}$$

Now solve the equation t=\frac{3±\sqrt{7}i}{2} when ± is minus. Subtract i\sqrt{7} from 3.

$$t=\frac{-\sqrt{7}i+3}{2}$$

The equation is now solved.

$$t=\frac{3+\sqrt{7}i}{2} t=\frac{-\sqrt{7}i+3}{2}$$

Steps for Completing the Square

Subtract 3 from -4 to get -7.

$$t^{2}-2t-3=-7+t$$

Subtract t from both sides.

$$t^{2}-2t-3-t=-7$$

Combine -2t and -t to get -3t.

$$t^{2}-3t-3=-7$$

Add 3 to both sides.

$$t^{2}-3t=-7+3$$

Add -7 and 3 to get -4.

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

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

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

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

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

Add -4 to \frac{9}{4}.

$$t^{2}-3t+\frac{9}{4}=-\frac{7}{4}$$

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

$$\left(t-\frac{3}{2}\right)^{2}=-\frac{7}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{7}{4}}$$

Simplify.

$$t-\frac{3}{2}=\frac{\sqrt{7}i}{2} t-\frac{3}{2}=-\frac{\sqrt{7}i}{2}$$

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

$$t=\frac{3+\sqrt{7}i}{2} t=\frac{-\sqrt{7}i+3}{2}$$