How do you solve #d^ { 2} - 3d = 88#?

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Answer 1 · 2017-05-12T11:36:41

#color(brown)(d=11# or #color(brown)(d=-8#

#d^2-3d=88#

#:.d^2-3d-88=0#

#:.=(d-11)(d+8)#

#:.d-11=0# or #d+8=0#

#:.color(brown)(d=11# or #color(brown)(d=-8#

substitute #color(brown)(d=11 #

#:.(11)^2-3(11)=88#

#:.121-33=88#

#:.color(brown)(88=88#

substitute#d=-8 #

#:.(-8)^2-3(-8)=88#

#:.64+24=88#

#:.color(brown)(88=88#