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find three consecutive odd integers if the sum of their squares is 56 more than three times the square of the smallest number.

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Jomastarviz
x² + (X+2)² + (X+4)² = 3X²+ 56
X² + (X²+4X+4) + (X²+8X+16) = 3X² + 56
 Remove those parentheses, then combine like terms.
3x² +12x + 20 = 3x² + 56        transpose 3x² to the right the 20 to the left, it will be:
12x = 36                finally, divide both sides by 12, the result will be
x =3