Solve the equation: 3a + 10 = a\(^2\)

  • A a = 5 or a = 2
  • B a = -5 or a = 2
  • C a = 10 or a = 0
  • D a = 5 or a = 0
  • E a = 5 or a = -2
waec 1991mathematics

The correct answer is E. a = 5 or a = -2

3a + 10 = a\(^2\)

a\(^2\) - 3a - 10 = 0

a\(^2\) - 5a + 2a - 10 = 0

a(a - 5) + 2(a - 5) = 0

(a - 5)(a + 2) = 0

a = 5 or a = -2.

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