Solve 2y^2-6y=3y-y^2

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

Solve for y

$$y=3$$

$$y=0$$

Steps Using Factoring

Subtract 3y from both sides.

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

Combine -6y and -3y to get -9y.

$$2y^{2}-9y=-y^{2}$$

Add y^{2} to both sides.

$$2y^{2}-9y+y^{2}=0$$

Combine 2y^{2} and y^{2} to get 3y^{2}.

$$3y^{2}-9y=0$$

Factor out y.

$$y\left(3y-9\right)=0$$

To find equation solutions, solve y=0 and 3y-9=0.

$$y=0 y=3$$

Steps Using the Quadratic Formula

Subtract 3y from both sides.

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

Combine -6y and -3y to get -9y.

$$2y^{2}-9y=-y^{2}$$

Add y^{2} to both sides.

$$2y^{2}-9y+y^{2}=0$$

Combine 2y^{2} and y^{2} to get 3y^{2}.

$$3y^{2}-9y=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -9 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of \left(-9\right)^{2}.

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

The opposite of -9 is 9.

$$y=\frac{9±9}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{9±9}{6}$$

Now solve the equation y=\frac{9±9}{6} when ± is plus. Add 9 to 9.

$$y=\frac{18}{6}$$

Divide 18 by 6.

$$y=3$$

Now solve the equation y=\frac{9±9}{6} when ± is minus. Subtract 9 from 9.

$$y=\frac{0}{6}$$

Divide 0 by 6.

$$y=0$$

The equation is now solved.

$$y=3 y=0$$

Steps for Completing the Square

Subtract 3y from both sides.

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

Combine -6y and -3y to get -9y.

$$2y^{2}-9y=-y^{2}$$

Add y^{2} to both sides.

$$2y^{2}-9y+y^{2}=0$$

Combine 2y^{2} and y^{2} to get 3y^{2}.

$$3y^{2}-9y=0$$

Divide both sides by 3.

$$\frac{3y^{2}-9y}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y^{2}+\left(-\frac{9}{3}\right)y=\frac{0}{3}$$

Divide -9 by 3.

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

Divide 0 by 3.

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

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.

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

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

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

Factor y^{2}-3y+\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}.

$$\left(y-\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$y-\frac{3}{2}=\frac{3}{2} y-\frac{3}{2}=-\frac{3}{2}$$

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

$$y=3 y=0$$