How do you solve # 4 x ^2 − 15 x = 3 #?

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Answer 1 · 2018-05-09T18:56:40

#x=(15+-sqrt273)/8#

#4x^2-15x=3#

#4x^2-15x-3=0#

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

with #a=4#, #b=-15#, #c=-3#

#x=(-(-15)+-sqrt((-15)^2-4(4)(-3)))/(2(4))=(15+-sqrt(225+48))/8=(15+-sqrt273)/8#

Answer 2 · 2018-05-09T19:05:45

#4x^2-15x=3#
#4x^2-15x-3=0#
#D= β^2-4αγ#
#D= 15^2-4*4*-3#
#D= 225+48#
#D= 273#

#x1,2=(-β+- sqrt273)/(2α)#
#x1,2=(15+- sqrt273)/8#
#x1=(15+ sqrt273)/8 #
# x2=(15- sqrt273)/8#