Solve y^2/4=3-y

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

Solve for y

$$y=-6$$

$$y=2$$

Steps Using Factoring

Multiply both sides of the equation by 4.

$$y^{2}=12-4y$$

Subtract 12 from both sides.

$$y^{2}-12=-4y$$

Add 4y to both sides.

$$y^{2}-12+4y=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$y^{2}+4y-12=0$$

To solve the equation, factor y^{2}+4y-12 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=-2 b=6$$

Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.

$$\left(y-2\right)\left(y+6\right)$$

To find equation solutions, solve y-2=0 and y+6=0.

$$y=2 y=-6$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 4.

$$y^{2}=12-4y$$

Subtract 12 from both sides.

$$y^{2}-12=-4y$$

Add 4y to both sides.

$$y^{2}-12+4y=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$y^{2}+4y-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-12. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-12\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=-2 b=6$$

Rewrite y^{2}+4y-12 as \left(y^{2}-2y\right)+\left(6y-12\right).

$$\left(y^{2}-2y\right)+\left(6y-12\right)$$

Factor out y in the first and 6 in the second group.

$$y\left(y-2\right)+6\left(y-2\right)$$

Factor out common term y-2 by using distributive property.

$$\left(y-2\right)\left(y+6\right)$$

To find equation solutions, solve y-2=0 and y+6=0.

$$y=2 y=-6$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$y^{2}=12-4y$$

Subtract 12 from both sides.

$$y^{2}-12=-4y$$

Add 4y to both sides.

$$y^{2}-12+4y=0$$

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.

$$y^{2}+4y-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-4±\sqrt{4^{2}-4\left(-12\right)}}{2}$$

Square 4.

$$y=\frac{-4±\sqrt{16-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$y=\frac{-4±\sqrt{16+48}}{2}$$

Add 16 to 48.

$$y=\frac{-4±\sqrt{64}}{2}$$

Take the square root of 64.

$$y=\frac{-4±8}{2}$$

Now solve the equation y=\frac{-4±8}{2} when ± is plus. Add -4 to 8.

$$y=\frac{4}{2}$$

Divide 4 by 2.

$$y=2$$

Now solve the equation y=\frac{-4±8}{2} when ± is minus. Subtract 8 from -4.

$$y=-\frac{12}{2}$$

Divide -12 by 2.

$$y=-6$$

The equation is now solved.

$$y=2 y=-6$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$y^{2}=12-4y$$

Add 4y to both sides.

$$y^{2}+4y=12$$

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

$$y^{2}+4y+2^{2}=12+2^{2}$$

Square 2.

$$y^{2}+4y+4=12+4$$

Add 12 to 4.

$$y^{2}+4y+4=16$$

Factor y^{2}+4y+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}.

$$\left(y+2\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+2\right)^{2}}=\sqrt{16}$$

Simplify.

$$y+2=4 y+2=-4$$

Subtract 2 from both sides of the equation.

$$y=2 y=-6$$