How do you solve #sqrt(3x+10)= 5-2x#?

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Answer 1 · 2016-04-11T20:25:50

#x_(1,2)={3/4,5}#

#sqrt(3x+10)=5-2x#

#(sqrt(3x+10))^2=(5-2x)^2#

#3x+10=(5-2x)^2#

#3x+10=25-20x+4x^2#

#4x^2-20x+25-3x-10=0#

#4x^2-23x+15=0#

#ax^2+bx+c=0#

#Delta=sqrt(b^2-4*a*c)#

#Delta=sqrt((-23)^2-4*4*15)#
#Delta=sqrt(529-240)#
#Delta=sqrt(289)#
#Delta=17#

#x_1=(-b-Delta)/(2*a)#

#x_1=(23-17)/(2*4)#

#x_1=6/8" "x_1=3/4#

#x_2=(-b+Delta)/(2*a)#

#x_2=(23+17)/(2*4)#

#x_2=20/8#

#x_2=5#

#x_(1,2)={3/4,5}#