How do you solve #5x^2-3x+3=-2x^2+3#?

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Answer 1 · 2017-01-23T14:55:57

#x=0 and x=+3/7#

Given:#" "5x^2-3x+3=-2x^2+3#

Spot that we have:#" "5x^2-3x+cancel(3)=-2x^2+cancel(3)#

Add #2x^2# to both sides

#5x^2+2x^2-3x=-2x^2+2x^2#

#7x^2-3x=0#

Factor out the #x#

#x(7x-3)=0#

#x=0# is 1 solution

The other solution is that #7x-3=0#

#=>7x=+3#

#=>x=+3/7#
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#color(blue)("Plotting "y=5x^2-3x+3 and )#
#color(blue)(y=-2x^2+3" gives the same result")#

Tony B