Sum of squares of two consecutive odd integers

The sum of the squares of two consecutive odd integers is 74. Find the two integers. 

x = 1st odd integer
x + 2 = 2nd odd integer

x² + (x + 2)² = 74 {sum of squares is 74}
x² + (x + 2)(x+ 2) = 74 {to square a binomial, multiply it by itself}
x² + x² + 4x + 4 = 74 {used foil method}
2x² + 4x + 4 = 74 {combined like terms}
2x² + 4x - 70 = 0 {subtracted 74 from both sides}
2(x² + 2x - 35) = 0 {factored 2 out}
x² + 2x - 35 = 0 {set in parentheses equal to 0, 2 cannot be 0}
(x + 7)(x - 5) = 0 {factored into two binomials}
x + 7 = 0 or x - 5 = 0 {set each factor equal to 0}
x = -7 or x = 5 {solved each equation for x}
x + 2 = -5 or x + 2 = 7 {substituted -7 and 5, in for x, into x + 2}

-7 and - 5 also 5 and 7 are the two consecutive odd integers


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