Solve 6y-2y^2

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

Factor

$$2y\left(3-y\right)$$

Solution Steps

Factor out 2.

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

Consider 3y-y^{2}. Factor out y.

$$y\left(3-y\right)$$

Rewrite the complete factored expression.

$$2y\left(-y+3\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

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

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.

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

Take the square root of 6^{2}.

$$y=\frac{-6±6}{2\left(-2\right)}$$

Multiply 2 times -2.

$$y=\frac{-6±6}{-4}$$

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

$$y=\frac{0}{-4}$$

Divide 0 by -4.

$$y=0$$

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

$$y=-\frac{12}{-4}$$

Divide -12 by -4.

$$y=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and 3 for x_{2}.

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

Evaluate

$$2y\left(3-y\right)$$