Solve 0=3t^2-4t+1

$$0 = 3 t ^ { 2 } - 4 t + 1$$

Solve for t

$$t=\frac{1}{3}\approx 0.333333333$$

$$t=1$$

Steps Using Factoring By Grouping

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

$$3t^{2}-4t+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3t^{2}+at+bt+1. To find a and b, set up a system to be solved.

$$a+b=-4 ab=3\times 1=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

Rewrite 3t^{2}-4t+1 as \left(3t^{2}-3t\right)+\left(-t+1\right).

$$\left(3t^{2}-3t\right)+\left(-t+1\right)$$

Factor out 3t in the first and -1 in the second group.

$$3t\left(t-1\right)-\left(t-1\right)$$

Factor out common term t-1 by using distributive property.

$$\left(t-1\right)\left(3t-1\right)$$

To find equation solutions, solve t-1=0 and 3t-1=0.

$$t=1 t=\frac{1}{3}$$

Steps Using the Quadratic Formula

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

$$3t^{2}-4t+1=0$$

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

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

Square -4.

$$t=\frac{-\left(-4\right)±\sqrt{16-4\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$t=\frac{-\left(-4\right)±\sqrt{16-12}}{2\times 3}$$

Add 16 to -12.

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

Take the square root of 4.

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

The opposite of -4 is 4.

$$t=\frac{4±2}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{4±2}{6}$$

Now solve the equation t=\frac{4±2}{6} when ± is plus. Add 4 to 2.

$$t=\frac{6}{6}$$

Divide 6 by 6.

$$t=1$$

Now solve the equation t=\frac{4±2}{6} when ± is minus. Subtract 2 from 4.

$$t=\frac{2}{6}$$

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

$$t=\frac{1}{3}$$

The equation is now solved.

$$t=1 t=\frac{1}{3}$$

Steps for Completing the Square

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

$$3t^{2}-4t+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$3t^{2}-4t=-1$$

Divide both sides by 3.

$$\frac{3t^{2}-4t}{3}=-\frac{1}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$t^{2}-\frac{4}{3}t=-\frac{1}{3}$$

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.

$$t^{2}-\frac{4}{3}t+\left(-\frac{2}{3}\right)^{2}=-\frac{1}{3}+\left(-\frac{2}{3}\right)^{2}$$

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

$$t^{2}-\frac{4}{3}t+\frac{4}{9}=-\frac{1}{3}+\frac{4}{9}$$

Add -\frac{1}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{4}{3}t+\frac{4}{9}=\frac{1}{9}$$

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}.

$$\left(t-\frac{2}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{2}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$t-\frac{2}{3}=\frac{1}{3} t-\frac{2}{3}=-\frac{1}{3}$$

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

$$t=1 t=\frac{1}{3}$$