Solve (2t-1)(t-3)=1

$$( 2 t - 1 ) ( t - 3 ) = 1$$

Solve for t

$$t = \frac{\sqrt{33} + 7}{4} \approx 3.186140662$$

$$t=\frac{7-\sqrt{33}}{4}\approx 0.313859338$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2t-1 by t-3 and combine like terms.

$$2t^{2}-7t+3=1$$

Subtract 1 from both sides.

$$2t^{2}-7t+3-1=0$$

Subtract 1 from 3 to get 2.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -7.

$$t=\frac{-\left(-7\right)±\sqrt{49-4\times 2\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$t=\frac{-\left(-7\right)±\sqrt{49-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$t=\frac{-\left(-7\right)±\sqrt{49-16}}{2\times 2}$$

Add 49 to -16.

$$t=\frac{-\left(-7\right)±\sqrt{33}}{2\times 2}$$

The opposite of -7 is 7.

$$t=\frac{7±\sqrt{33}}{2\times 2}$$

Multiply 2 times 2.

$$t=\frac{7±\sqrt{33}}{4}$$

Now solve the equation t=\frac{7±\sqrt{33}}{4} when ± is plus. Add 7 to \sqrt{33}.

$$t=\frac{\sqrt{33}+7}{4}$$

Now solve the equation t=\frac{7±\sqrt{33}}{4} when ± is minus. Subtract \sqrt{33} from 7.

$$t=\frac{7-\sqrt{33}}{4}$$

The equation is now solved.

$$t=\frac{\sqrt{33}+7}{4} t=\frac{7-\sqrt{33}}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 2t-1 by t-3 and combine like terms.

$$2t^{2}-7t+3=1$$

Subtract 3 from both sides.

$$2t^{2}-7t=1-3$$

Subtract 3 from 1 to get -2.

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

Divide both sides by 2.

$$\frac{2t^{2}-7t}{2}=-\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$t^{2}-\frac{7}{2}t=-\frac{2}{2}$$

Divide -2 by 2.

$$t^{2}-\frac{7}{2}t=-1$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{7}{2}t+\left(-\frac{7}{4}\right)^{2}=-1+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{7}{2}t+\frac{49}{16}=-1+\frac{49}{16}$$

Add -1 to \frac{49}{16}.

$$t^{2}-\frac{7}{2}t+\frac{49}{16}=\frac{33}{16}$$

Factor t^{2}-\frac{7}{2}t+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{7}{4}\right)^{2}=\frac{33}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{7}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Simplify.

$$t-\frac{7}{4}=\frac{\sqrt{33}}{4} t-\frac{7}{4}=-\frac{\sqrt{33}}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$t=\frac{\sqrt{33}+7}{4} t=\frac{7-\sqrt{33}}{4}$$