Solve 2y^2-12y+17=(y-3)^2

$$2 y ^ { 2 } - 12 y + 17 = ( y - 3 ) ^ { 2 }$$

Solve for y

$$y=2$$

$$y=4$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-3\right)^{2}.

$$2y^{2}-12y+17=y^{2}-6y+9$$

Subtract y^{2} from both sides.

$$2y^{2}-12y+17-y^{2}=-6y+9$$

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

$$y^{2}-12y+17=-6y+9$$

Add 6y to both sides.

$$y^{2}-12y+17+6y=9$$

Combine -12y and 6y to get -6y.

$$y^{2}-6y+17=9$$

Subtract 9 from both sides.

$$y^{2}-6y+17-9=0$$

Subtract 9 from 17 to get 8.

$$y^{2}-6y+8=0$$

To solve the equation, factor y^{2}-6y+8 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.

$$a+b=-6 ab=8$$

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 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-4 b=-2$$

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

$$\left(y-4\right)\left(y-2\right)$$

To find equation solutions, solve y-4=0 and y-2=0.

$$y=4 y=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-3\right)^{2}.

$$2y^{2}-12y+17=y^{2}-6y+9$$

Subtract y^{2} from both sides.

$$2y^{2}-12y+17-y^{2}=-6y+9$$

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

$$y^{2}-12y+17=-6y+9$$

Add 6y to both sides.

$$y^{2}-12y+17+6y=9$$

Combine -12y and 6y to get -6y.

$$y^{2}-6y+17=9$$

Subtract 9 from both sides.

$$y^{2}-6y+17-9=0$$

Subtract 9 from 17 to get 8.

$$y^{2}-6y+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+8. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 8=8$$

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 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-4 b=-2$$

Rewrite y^{2}-6y+8 as \left(y^{2}-4y\right)+\left(-2y+8\right).

$$\left(y^{2}-4y\right)+\left(-2y+8\right)$$

Factor out y in the first and -2 in the second group.

$$y\left(y-4\right)-2\left(y-4\right)$$

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

$$\left(y-4\right)\left(y-2\right)$$

To find equation solutions, solve y-4=0 and y-2=0.

$$y=4 y=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-3\right)^{2}.

$$2y^{2}-12y+17=y^{2}-6y+9$$

Subtract y^{2} from both sides.

$$2y^{2}-12y+17-y^{2}=-6y+9$$

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

$$y^{2}-12y+17=-6y+9$$

Add 6y to both sides.

$$y^{2}-12y+17+6y=9$$

Combine -12y and 6y to get -6y.

$$y^{2}-6y+17=9$$

Subtract 9 from both sides.

$$y^{2}-6y+17-9=0$$

Subtract 9 from 17 to get 8.

$$y^{2}-6y+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8}}{2}$$

Square -6.

$$y=\frac{-\left(-6\right)±\sqrt{36-4\times 8}}{2}$$

Multiply -4 times 8.

$$y=\frac{-\left(-6\right)±\sqrt{36-32}}{2}$$

Add 36 to -32.

$$y=\frac{-\left(-6\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$y=\frac{-\left(-6\right)±2}{2}$$

The opposite of -6 is 6.

$$y=\frac{6±2}{2}$$

Now solve the equation y=\frac{6±2}{2} when ± is plus. Add 6 to 2.

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

Divide 8 by 2.

$$y=4$$

Now solve the equation y=\frac{6±2}{2} when ± is minus. Subtract 2 from 6.

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

Divide 4 by 2.

$$y=2$$

The equation is now solved.

$$y=4 y=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-3\right)^{2}.

$$2y^{2}-12y+17=y^{2}-6y+9$$

Subtract y^{2} from both sides.

$$2y^{2}-12y+17-y^{2}=-6y+9$$

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

$$y^{2}-12y+17=-6y+9$$

Add 6y to both sides.

$$y^{2}-12y+17+6y=9$$

Combine -12y and 6y to get -6y.

$$y^{2}-6y+17=9$$

Subtract 17 from both sides.

$$y^{2}-6y=9-17$$

Subtract 17 from 9 to get -8.

$$y^{2}-6y=-8$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-6y+\left(-3\right)^{2}=-8+\left(-3\right)^{2}$$

Square -3.

$$y^{2}-6y+9=-8+9$$

Add -8 to 9.

$$y^{2}-6y+9=1$$

Factor y^{2}-6y+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-3\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-3\right)^{2}}=\sqrt{1}$$

Simplify.

$$y-3=1 y-3=-1$$

Add 3 to both sides of the equation.

$$y=4 y=2$$