Solve y^2=16(5-y)

$${ y }^{ 2 } =16(5-y)$$

Solve for y

$$y=-20$$

$$y=4$$

Steps Using Factoring

Use the distributive property to multiply 16 by 5-y.

$$y^{2}=80-16y$$

Subtract 80 from both sides.

$$y^{2}-80=-16y$$

Add 16y to both sides.

$$y^{2}-80+16y=0$$

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

$$y^{2}+16y-80=0$$

To solve the equation, factor y^{2}+16y-80 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.

$$a+b=16 ab=-80$$

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 -80.

$$-1,80 -2,40 -4,20 -5,16 -8,10$$

Calculate the sum for each pair.

$$-1+80=79 -2+40=38 -4+20=16 -5+16=11 -8+10=2$$

The solution is the pair that gives sum 16.

$$a=-4 b=20$$

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

$$\left(y-4\right)\left(y+20\right)$$

To find equation solutions, solve y-4=0 and y+20=0.

$$y=4 y=-20$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 16 by 5-y.

$$y^{2}=80-16y$$

Subtract 80 from both sides.

$$y^{2}-80=-16y$$

Add 16y to both sides.

$$y^{2}-80+16y=0$$

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

$$y^{2}+16y-80=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-80. To find a and b, set up a system to be solved.

$$a+b=16 ab=1\left(-80\right)=-80$$

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 -80.

$$-1,80 -2,40 -4,20 -5,16 -8,10$$

Calculate the sum for each pair.

$$-1+80=79 -2+40=38 -4+20=16 -5+16=11 -8+10=2$$

The solution is the pair that gives sum 16.

$$a=-4 b=20$$

Rewrite y^{2}+16y-80 as \left(y^{2}-4y\right)+\left(20y-80\right).

$$\left(y^{2}-4y\right)+\left(20y-80\right)$$

Factor out y in the first and 20 in the second group.

$$y\left(y-4\right)+20\left(y-4\right)$$

Factor out common term y-4 by using distributive property.

$$\left(y-4\right)\left(y+20\right)$$

To find equation solutions, solve y-4=0 and y+20=0.

$$y=4 y=-20$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 16 by 5-y.

$$y^{2}=80-16y$$

Subtract 80 from both sides.

$$y^{2}-80=-16y$$

Add 16y to both sides.

$$y^{2}-80+16y=0$$

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.

$$y^{2}+16y-80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 16 for b, and -80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-16±\sqrt{16^{2}-4\left(-80\right)}}{2}$$

Square 16.

$$y=\frac{-16±\sqrt{256-4\left(-80\right)}}{2}$$

Multiply -4 times -80.

$$y=\frac{-16±\sqrt{256+320}}{2}$$

Add 256 to 320.

$$y=\frac{-16±\sqrt{576}}{2}$$

Take the square root of 576.

$$y=\frac{-16±24}{2}$$

Now solve the equation y=\frac{-16±24}{2} when ± is plus. Add -16 to 24.

$$y=\frac{8}{2}$$

Divide 8 by 2.

$$y=4$$

Now solve the equation y=\frac{-16±24}{2} when ± is minus. Subtract 24 from -16.

$$y=-\frac{40}{2}$$

Divide -40 by 2.

$$y=-20$$

The equation is now solved.

$$y=4 y=-20$$

Steps for Completing the Square

Use the distributive property to multiply 16 by 5-y.

$$y^{2}=80-16y$$

Add 16y to both sides.

$$y^{2}+16y=80$$

Divide 16, the coefficient of the x term, by 2 to get 8. Then add the square of 8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+16y+8^{2}=80+8^{2}$$

Square 8.

$$y^{2}+16y+64=80+64$$

Add 80 to 64.

$$y^{2}+16y+64=144$$

Factor y^{2}+16y+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+8\right)^{2}=144$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+8\right)^{2}}=\sqrt{144}$$

Simplify.

$$y+8=12 y+8=-12$$

Subtract 8 from both sides of the equation.

$$y=4 y=-20$$