Solve 6y^2=3y

$$6 y ^ { 2 } = 3 y$$

Solve for y

$$y=\frac{1}{2}=0.5$$

$$y=0$$

Steps Using Factoring

Subtract 3y from both sides.

$$6y^{2}-3y=0$$

Factor out y.

$$y\left(6y-3\right)=0$$

To find equation solutions, solve y=0 and 6y-3=0.

$$y=0 y=\frac{1}{2}$$

Steps Using the Quadratic Formula

Subtract 3y from both sides.

$$6y^{2}-3y=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}}}{2\times 6}$$

Take the square root of \left(-3\right)^{2}.

$$y=\frac{-\left(-3\right)±3}{2\times 6}$$

The opposite of -3 is 3.

$$y=\frac{3±3}{2\times 6}$$

Multiply 2 times 6.

$$y=\frac{3±3}{12}$$

Now solve the equation y=\frac{3±3}{12} when ± is plus. Add 3 to 3.

$$y=\frac{6}{12}$$

Reduce the fraction \frac{6}{12} to lowest terms by extracting and canceling out 6.

$$y=\frac{1}{2}$$

Now solve the equation y=\frac{3±3}{12} when ± is minus. Subtract 3 from 3.

$$y=\frac{0}{12}$$

Divide 0 by 12.

$$y=0$$

The equation is now solved.

$$y=\frac{1}{2} y=0$$

Steps for Completing the Square

Subtract 3y from both sides.

$$6y^{2}-3y=0$$

Divide both sides by 6.

$$\frac{6y^{2}-3y}{6}=\frac{0}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$y^{2}+\left(-\frac{3}{6}\right)y=\frac{0}{6}$$

Reduce the fraction \frac{-3}{6} to lowest terms by extracting and canceling out 3.

$$y^{2}-\frac{1}{2}y=\frac{0}{6}$$

Divide 0 by 6.

$$y^{2}-\frac{1}{2}y=0$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{1}{2}y+\frac{1}{16}=\frac{1}{16}$$

Factor y^{2}-\frac{1}{2}y+\frac{1}{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}.

$$\left(y-\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$y-\frac{1}{4}=\frac{1}{4} y-\frac{1}{4}=-\frac{1}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$y=\frac{1}{2} y=0$$