Find two consecutive odd integers such that the sum of their squares is 130.
x = 1st odd integer
x + 2 = 2nd odd integer {odd integers increase by two each time}
x² + (x + 2)² = 130 {sum of their squares is 130}
x² + (x + 2)(x + 2) = 130 {when you square a binomial, multiply it by itself}
x² + x² + 4x + 4 = 130 {used foil method}
2x² + 4x + 4 = 130 {combined like terms}
2x² + 4x - 126 = 0 {subtracted 130 from both sides}
2(x² + 2x - 63) = 0 {factored 2 out}
2(x + 9)(x - 7) = 0 {factored into two binomials}
x + 9 = 0 or x - 7 = 0 {set each factor equal to 0, excluding the 2}
x = -9 or x = 7 {solved each equation for x}
x + 2 = -7 or x + 2 = 9 {substituted -9 and 7, in for x, into x + 2}
-9 and -7, also 7 and 9 are the two consecutive odd integers
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