Solve 2y^2-10y+16=(y-1)^2

$$2 y ^ { 2 } - 10 y + 16 = ( y - 1 ) ^ { 2 }$$

Solve for y

$$y=3$$

$$y=5$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-1\right)^{2}.

$$2y^{2}-10y+16=y^{2}-2y+1$$

Subtract y^{2} from both sides.

$$2y^{2}-10y+16-y^{2}=-2y+1$$

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

$$y^{2}-10y+16=-2y+1$$

Add 2y to both sides.

$$y^{2}-10y+16+2y=1$$

Combine -10y and 2y to get -8y.

$$y^{2}-8y+16=1$$

Subtract 1 from both sides.

$$y^{2}-8y+16-1=0$$

Subtract 1 from 16 to get 15.

$$y^{2}-8y+15=0$$

To solve the equation, factor y^{2}-8y+15 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.

$$a+b=-8 ab=15$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-5 b=-3$$

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

$$\left(y-5\right)\left(y-3\right)$$

To find equation solutions, solve y-5=0 and y-3=0.

$$y=5 y=3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-1\right)^{2}.

$$2y^{2}-10y+16=y^{2}-2y+1$$

Subtract y^{2} from both sides.

$$2y^{2}-10y+16-y^{2}=-2y+1$$

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

$$y^{2}-10y+16=-2y+1$$

Add 2y to both sides.

$$y^{2}-10y+16+2y=1$$

Combine -10y and 2y to get -8y.

$$y^{2}-8y+16=1$$

Subtract 1 from both sides.

$$y^{2}-8y+16-1=0$$

Subtract 1 from 16 to get 15.

$$y^{2}-8y+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+15. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 15=15$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-5 b=-3$$

Rewrite y^{2}-8y+15 as \left(y^{2}-5y\right)+\left(-3y+15\right).

$$\left(y^{2}-5y\right)+\left(-3y+15\right)$$

Factor out y in the first and -3 in the second group.

$$y\left(y-5\right)-3\left(y-5\right)$$

Factor out common term y-5 by using distributive property.

$$\left(y-5\right)\left(y-3\right)$$

To find equation solutions, solve y-5=0 and y-3=0.

$$y=5 y=3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-1\right)^{2}.

$$2y^{2}-10y+16=y^{2}-2y+1$$

Subtract y^{2} from both sides.

$$2y^{2}-10y+16-y^{2}=-2y+1$$

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

$$y^{2}-10y+16=-2y+1$$

Add 2y to both sides.

$$y^{2}-10y+16+2y=1$$

Combine -10y and 2y to get -8y.

$$y^{2}-8y+16=1$$

Subtract 1 from both sides.

$$y^{2}-8y+16-1=0$$

Subtract 1 from 16 to get 15.

$$y^{2}-8y+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 15}}{2}$$

Square -8.

$$y=\frac{-\left(-8\right)±\sqrt{64-4\times 15}}{2}$$

Multiply -4 times 15.

$$y=\frac{-\left(-8\right)±\sqrt{64-60}}{2}$$

Add 64 to -60.

$$y=\frac{-\left(-8\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$y=\frac{-\left(-8\right)±2}{2}$$

The opposite of -8 is 8.

$$y=\frac{8±2}{2}$$

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

$$y=\frac{10}{2}$$

Divide 10 by 2.

$$y=5$$

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

$$y=\frac{6}{2}$$

Divide 6 by 2.

$$y=3$$

The equation is now solved.

$$y=5 y=3$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-1\right)^{2}.

$$2y^{2}-10y+16=y^{2}-2y+1$$

Subtract y^{2} from both sides.

$$2y^{2}-10y+16-y^{2}=-2y+1$$

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

$$y^{2}-10y+16=-2y+1$$

Add 2y to both sides.

$$y^{2}-10y+16+2y=1$$

Combine -10y and 2y to get -8y.

$$y^{2}-8y+16=1$$

Factor y^{2}-8y+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$y-4=1 y-4=-1$$

Add 4 to both sides of the equation.

$$y=5 y=3$$