Algebra 1: Common Core (15th Edition)

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.