Solve (y-2)^2-7(y-2)+6=0

$$( y - 2 ) ^ { 2 } - 7 ( y - 2 ) + 6 = 0$$

Solve for y

$$y=3$$

$$y=8$$

Steps Using Factoring

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

$$y^{2}-4y+4-7\left(y-2\right)+6=0$$

Use the distributive property to multiply -7 by y-2.

$$y^{2}-4y+4-7y+14+6=0$$

Combine -4y and -7y to get -11y.

$$y^{2}-11y+4+14+6=0$$

Add 4 and 14 to get 18.

$$y^{2}-11y+18+6=0$$

Add 18 and 6 to get 24.

$$y^{2}-11y+24=0$$

To solve the equation, factor y^{2}-11y+24 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.

$$a+b=-11 ab=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -11.

$$a=-8 b=-3$$

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

$$\left(y-8\right)\left(y-3\right)$$

To find equation solutions, solve y-8=0 and y-3=0.

$$y=8 y=3$$

Steps Using Factoring By Grouping

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

$$y^{2}-4y+4-7\left(y-2\right)+6=0$$

Use the distributive property to multiply -7 by y-2.

$$y^{2}-4y+4-7y+14+6=0$$

Combine -4y and -7y to get -11y.

$$y^{2}-11y+4+14+6=0$$

Add 4 and 14 to get 18.

$$y^{2}-11y+18+6=0$$

Add 18 and 6 to get 24.

$$y^{2}-11y+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+24. To find a and b, set up a system to be solved.

$$a+b=-11 ab=1\times 24=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -11.

$$a=-8 b=-3$$

Rewrite y^{2}-11y+24 as \left(y^{2}-8y\right)+\left(-3y+24\right).

$$\left(y^{2}-8y\right)+\left(-3y+24\right)$$

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

$$y\left(y-8\right)-3\left(y-8\right)$$

Factor out common term y-8 by using distributive property.

$$\left(y-8\right)\left(y-3\right)$$

To find equation solutions, solve y-8=0 and y-3=0.

$$y=8 y=3$$

Steps Using the Quadratic Formula

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

$$y^{2}-4y+4-7\left(y-2\right)+6=0$$

Use the distributive property to multiply -7 by y-2.

$$y^{2}-4y+4-7y+14+6=0$$

Combine -4y and -7y to get -11y.

$$y^{2}-11y+4+14+6=0$$

Add 4 and 14 to get 18.

$$y^{2}-11y+18+6=0$$

Add 18 and 6 to get 24.

$$y^{2}-11y+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -11 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 24}}{2}$$

Square -11.

$$y=\frac{-\left(-11\right)±\sqrt{121-4\times 24}}{2}$$

Multiply -4 times 24.

$$y=\frac{-\left(-11\right)±\sqrt{121-96}}{2}$$

Add 121 to -96.

$$y=\frac{-\left(-11\right)±\sqrt{25}}{2}$$

Take the square root of 25.

$$y=\frac{-\left(-11\right)±5}{2}$$

The opposite of -11 is 11.

$$y=\frac{11±5}{2}$$

Now solve the equation y=\frac{11±5}{2} when ± is plus. Add 11 to 5.

$$y=\frac{16}{2}$$

Divide 16 by 2.

$$y=8$$

Now solve the equation y=\frac{11±5}{2} when ± is minus. Subtract 5 from 11.

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

Divide 6 by 2.

$$y=3$$

The equation is now solved.

$$y=8 y=3$$

Steps for Completing the Square

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

$$y^{2}-4y+4-7\left(y-2\right)+6=0$$

Use the distributive property to multiply -7 by y-2.

$$y^{2}-4y+4-7y+14+6=0$$

Combine -4y and -7y to get -11y.

$$y^{2}-11y+4+14+6=0$$

Add 4 and 14 to get 18.

$$y^{2}-11y+18+6=0$$

Add 18 and 6 to get 24.

$$y^{2}-11y+24=0$$

Subtract 24 from both sides. Anything subtracted from zero gives its negation.

$$y^{2}-11y=-24$$

Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-11y+\left(-\frac{11}{2}\right)^{2}=-24+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-11y+\frac{121}{4}=-24+\frac{121}{4}$$

Add -24 to \frac{121}{4}.

$$y^{2}-11y+\frac{121}{4}=\frac{25}{4}$$

Factor y^{2}-11y+\frac{121}{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}.

$$\left(y-\frac{11}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{11}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$y-\frac{11}{2}=\frac{5}{2} y-\frac{11}{2}=-\frac{5}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$y=8 y=3$$