How do you solve #(2m + 3)(4m + 3) = 0#?

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Answer 1 · 2018-03-14T11:39:52

#m=-12, -6#

Multiply the equation using FOIL

#( 2m + 3 ) ( 4m + 3 ) = 0#
#(2m * 4m) + (2m * 3) + (3 * 4m) + (3 * 3) = 0#
#8m^2 + 6m + 12m + 9 = 0#
#8m^2 + 18m + 9 = 0#

Since it would be sort of difficult to factor mentally,
we use the quadratic equation
(or just keep plugging in factors till you get the answer)

#(- b +- sqrt(b^2 - 4ac))/(2a)#

#(-(18) +- sqrt((18)^2 - 4(8)(9)))/(2(8))#

#(-18 +- sqrt(324-288))/16#

#(-18 +- sqrt(36))/16#

#(-18+-6)/16#

#(-9+-3)/8=0#

#m=-12, -6#