Chapter 9 - Quadratic Functions and Equations - 9-3 Solving Quadratic Equations - Got It? - Page 562: 2

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a) $m^2-36=0$ $m^2=36$ $m=\pm \sqrt 36$ $m = \pm 6$ There are two solutions, $ \pm 6$ b) $3x^2+15=0$ $3x^2=-15$ $x^2=-5$ There is no real-number solution. c) $4d^2+16=16$ $4d^2=0$ $d^2=0$ $d=0$ There is one solution, 0.