Solve y-6y+2y^2-(3y-y^2)=0

$$y - 6 y + 2 y ^ { 2 } - ( 3 y - y ^ { 2 } ) = 0$$

Solve for y

$$y = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

$$y=0$$

Steps Using Factoring

Combine y and -6y to get -5y.

$$-5y+2y^{2}-\left(3y-y^{2}\right)=0$$

To find the opposite of 3y-y^{2}, find the opposite of each term.

$$-5y+2y^{2}-3y+y^{2}=0$$

Combine -5y and -3y to get -8y.

$$-8y+2y^{2}+y^{2}=0$$

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

$$-8y+3y^{2}=0$$

Factor out y.

$$y\left(-8+3y\right)=0$$

To find equation solutions, solve y=0 and -8+3y=0.

$$y=0 y=\frac{8}{3}$$

Steps Using the Quadratic Formula

Combine y and -6y to get -5y.

$$-5y+2y^{2}-\left(3y-y^{2}\right)=0$$

To find the opposite of 3y-y^{2}, find the opposite of each term.

$$-5y+2y^{2}-3y+y^{2}=0$$

Combine -5y and -3y to get -8y.

$$-8y+2y^{2}+y^{2}=0$$

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

$$-8y+3y^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$3y^{2}-8y=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -8 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2\times 3}$$

Take the square root of \left(-8\right)^{2}.

$$y=\frac{-\left(-8\right)±8}{2\times 3}$$

The opposite of -8 is 8.

$$y=\frac{8±8}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{8±8}{6}$$

Now solve the equation y=\frac{8±8}{6} when ± is plus. Add 8 to 8.

$$y=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

$$y=\frac{8}{3}$$

Now solve the equation y=\frac{8±8}{6} when ± is minus. Subtract 8 from 8.

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

Divide 0 by 6.

$$y=0$$

The equation is now solved.

$$y=\frac{8}{3} y=0$$

Steps for Completing the Square

Combine y and -6y to get -5y.

$$-5y+2y^{2}-\left(3y-y^{2}\right)=0$$

To find the opposite of 3y-y^{2}, find the opposite of each term.

$$-5y+2y^{2}-3y+y^{2}=0$$

Combine -5y and -3y to get -8y.

$$-8y+2y^{2}+y^{2}=0$$

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

$$-8y+3y^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$3y^{2}-8y=0$$

Divide both sides by 3.

$$\frac{3y^{2}-8y}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y^{2}-\frac{8}{3}y=\frac{0}{3}$$

Divide 0 by 3.

$$y^{2}-\frac{8}{3}y=0$$

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{8}{3}y+\left(-\frac{4}{3}\right)^{2}=\left(-\frac{4}{3}\right)^{2}$$

Square -\frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{8}{3}y+\frac{16}{9}=\frac{16}{9}$$

Factor y^{2}-\frac{8}{3}y+\frac{16}{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}.

$$\left(y-\frac{4}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{4}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$y-\frac{4}{3}=\frac{4}{3} y-\frac{4}{3}=-\frac{4}{3}$$

Add \frac{4}{3} to both sides of the equation.

$$y=\frac{8}{3} y=0$$