Solve for y (y+1)^2-20(y+1)+64=0

${(y + 1)}^{2} - 20 (y + 1) + 64 = 0$

Factor the left side of the equation.

Let $u = y + 1$ . Substitute $u$ for all occurrences of $y + 1$ .

$u^{2} - 20 u + 64 = 0$

Factor $u^{2} - 20 u + 64$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $64$ and whose sum is $- 20$ .

$- 16, - 4$

Write the factored form using these integers.

$(u - 16) (u - 4) = 0$

Replace all occurrences of $u$ with $y + 1$ .

$(y + 1 - 16) (y + 1 - 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y + 1 - 16 = 0$

$y + 1 - 4 = 0$

Set $y + 1 - 16$ equal to $0$ and solve for $y$ .

Set $y + 1 - 16$ equal to $0$ .

$y + 1 - 16 = 0$

Solve $y + 1 - 16 = 0$ for $y$ .

Subtract $16$ from $1$ .

$y - 15 = 0$

Add $15$ to both sides of the equation.

$y = 15$

Set $y + 1 - 4$ equal to $0$ and solve for $y$ .

Set $y + 1 - 4$ equal to $0$ .

$y + 1 - 4 = 0$

Solve $y + 1 - 4 = 0$ for $y$ .

Subtract $4$ from $1$ .

$y - 3 = 0$

Add $3$ to both sides of the equation.

$y = 3$

The final solution is all the values that make $(y + 1 - 16) (y + 1 - 4) = 0$ true.

$y = 15, 3$

Solve for y (y+1)^2-20(y+1)+64=0

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