How do you factor #2d^2 + 7d - 15#?

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Answer 1 · 2015-06-04T22:04:36

#2d^2+7d-15 = (d+5)(2d-3)#

I used the quadratic formula to solve this one.

#a=2#, #b=7#, #c=-15#

#Delta = b^2-4ac = 49+120 = 169 = 13^2#

So #2d^2+7d-15 = 0# has roots:

#d = (-b+-sqrt(Delta))/(2a)#

#=(-7+-13)/4#

That is #d = -5# or #d = 3/2#

Hence factors #(d+5)# and #(2d-3)#, respectively.