APPROACH 1: Plug in each answer choice to see which value does NOT satisfy the equation (slow but...)
For example, C) x = 3
We get: (3² + 3 - 20)² - 2(3² + 3 - 20) - 63 = 17
Evaluate: (-8)² - 2(-8) - 63 = 17
Simplify: 64 - 16 - 63 = 17
Works! So, x = 3 is a valid solution

Try D) x = 4
We get: (4² + 4 - 20)² - 2(4² + 4 - 20) - 63 = 17
Evaluate: (0)² - 2(0) - 63 = 17
Simplify: -63 = 17
Works! So, x = 4 is NOT a valid solution

Answer: D
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APPROACH 2: Let k = x² + x - 20
Now replace x² + x - 20 with k to get: k² - 2k - 63 = 17
Subtract 17 from both sides to get: k² - 2k - 80 = 0
Factor: (k - 10)(k + 8) = 0
So, either k = 10 or k = -8

Now replace k with x² + x - 20 to get:
x² + x - 20 = 10 and x² + x - 20 = -8

Take: x² + x - 20 = 10
Rearrange: x² + x - 30 = 0
Factor: (x + 6)(x - 5) = 0
Solutions: x = -6 and x = 5

Take: x² + x - 20 = -8
Rearrange: x² + x - 12 = 0
Factor: (x + 4)(x - 3) = 0
Solutions: x = -4 and x = 3

ALL SOLUTIONS: x = -6, 5, -4 and 3

Answer: D
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APPROACH 3: Examine: x² + x - 20
Factor to get: (x + 5)(x - 4)
Notice that, when x = 4, (x + 5)(x - 4) = (4 + 5)(4 - 4) = 0

So, when x = 4, we get: (0)² - 2(0) - 63 = 17
When we simplify, we get: -63 = 17
In other words, x = 4 is NOT a solution to the original equation.

Answer: D

Cheers,
Brent