Find the whole number such that twice its square is 11 more than 21 times the number.​

Let x = the whole number

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From: "twice its square is 11 more than 21 times the original number"

2x^2 = 21x+11

2x^2 - 21x - 11 = 0

Factoring the left:

(2x+1)(x-11) = 0

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Setting each term on the left to zero:

(2x+1) = 0

2x = -1

x = -1/2

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(x-11) = 0

x = 11

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Solution set:

x = {-1/2, 11}

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Since the problem asks for a "whole number" we eliminate the -1/2 leaving us with:

x = 11