Solve (4-y)=(2-y)(2+y)

$$( 4 - y ) = ( 2 - y ) ( 2 + y )$$

Solve for y

$$y=1$$

$$y=0$$

Steps Using the Quadratic Formula

Consider \left(2-y\right)\left(2+y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$4-y=4-y^{2}$$

Subtract 4 from both sides.

$$4-y-4=-y^{2}$$

Subtract 4 from 4 to get 0.

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

Add y^{2} to both sides.

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

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.

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

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

$$y=\frac{-\left(-1\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$y=\frac{-\left(-1\right)±1}{2}$$

The opposite of -1 is 1.

$$y=\frac{1±1}{2}$$

Now solve the equation y=\frac{1±1}{2} when ± is plus. Add 1 to 1.

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

Divide 2 by 2.

$$y=1$$

Now solve the equation y=\frac{1±1}{2} when ± is minus. Subtract 1 from 1.

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

Divide 0 by 2.

$$y=0$$

The equation is now solved.

$$y=1 y=0$$

Steps for Completing the Square

Consider \left(2-y\right)\left(2+y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$4-y=4-y^{2}$$

Add y^{2} to both sides.

$$4-y+y^{2}=4$$

Subtract 4 from both sides.

$$-y+y^{2}=4-4$$

Subtract 4 from 4 to get 0.

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

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.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-y+\frac{1}{4}=\frac{1}{4}$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$y=1 y=0$$