(z+1)(z-12)=11(z-3)

Solution for (z+1)(z-12)=11(z-3) equation:

Simplifying
(z + 1)(z + -12) = 11(z + -3)

Reorder the terms:
(1 + z)(z + -12) = 11(z + -3)

Reorder the terms:
(1 + z)(-12 + z) = 11(z + -3)

Multiply (1 + z) * (-12 + z)
(1(-12 + z) + z(-12 + z)) = 11(z + -3)
((-12 * 1 + z * 1) + z(-12 + z)) = 11(z + -3)
((-12 + 1z) + z(-12 + z)) = 11(z + -3)
(-12 + 1z + (-12 * z + z * z)) = 11(z + -3)
(-12 + 1z + (-12z + z2)) = 11(z + -3)

Combine like terms: 1z + -12z = -11z
(-12 + -11z + z2) = 11(z + -3)

Reorder the terms:
-12 + -11z + z2 = 11(-3 + z)
-12 + -11z + z2 = (-3 * 11 + z * 11)
-12 + -11z + z2 = (-33 + 11z)

Solving
-12 + -11z + z2 = -33 + 11z

Solving for variable 'z'.

Reorder the terms:
-12 + 33 + -11z + -11z + z2 = -33 + 11z + 33 + -11z

Combine like terms: -12 + 33 = 21
21 + -11z + -11z + z2 = -33 + 11z + 33 + -11z

Combine like terms: -11z + -11z = -22z
21 + -22z + z2 = -33 + 11z + 33 + -11z

Reorder the terms:
21 + -22z + z2 = -33 + 33 + 11z + -11z

Combine like terms: -33 + 33 = 0
21 + -22z + z2 = 0 + 11z + -11z
21 + -22z + z2 = 11z + -11z

Combine like terms: 11z + -11z = 0
21 + -22z + z2 = 0

Factor a trinomial.
(1 + -1z)(21 + -1z) = 0

Subproblem 1
Set the factor '(1 + -1z)' equal to zero and attempt to solve:

Simplifying
1 + -1z = 0

Solving
1 + -1z = 0

Move all terms containing z to the left, all other terms to the right.

Add '-1' to each side of the equation.
1 + -1 + -1z = 0 + -1

Combine like terms: 1 + -1 = 0
0 + -1z = 0 + -1
-1z = 0 + -1

Combine like terms: 0 + -1 = -1
-1z = -1

Divide each side by '-1'.
z = 1

Simplifying
z = 1
Subproblem 2
Set the factor '(21 + -1z)' equal to zero and attempt to solve:

Simplifying
21 + -1z = 0

Solving
21 + -1z = 0

Move all terms containing z to the left, all other terms to the right.

Add '-21' to each side of the equation.
21 + -21 + -1z = 0 + -21

Combine like terms: 21 + -21 = 0
0 + -1z = 0 + -21
-1z = 0 + -21

Combine like terms: 0 + -21 = -21
-1z = -21

Divide each side by '-1'.
z = 21

Simplifying
z = 21
Solution
z = {1, 21}