How do you factor #2d ^ { 2} + 3d - 20= 0#?

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Answer 1 · 2017-08-09T01:39:08

The factors are #(2d-5)# and #(d+4)#.

If you need to solve it, then #d=2 1/2# or #d=-4#

#2d^2+3d-20=0#

Factorise #2xx20# to get two numbers with a difference of #3#.

#40=>8*5#

#2d^2+8d-5d-20=0#

#2d(d+4)-5(d+4)=0#

#(2d-5)(d+4)=0#

If you need to go further and solve this, then:

#2d-5=0# and #d+4=0#

#d=5/2=2 1/2# or #d=-4#