Question #85932

Question

726 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-08T19:24:18

#x=-1 # and # x=-4/5#

#5x^2+9x=-4#

#:.5x+9x+4=0#

#:.(x+1)(5x+4)#

#:.x+1=0# and #5x+4=0#

#:.color(blue)(x=-1# and #5x=-4#

#:.color(blue)(x=-1# and #color(blue)(x=-4/5#

substitute #color(blue)(x=-1#

#:.5(color(blue)(-1))^2+9(color(blue)(-1))=-4#

#:.5(color(blue)(-1))(color(blue)(-1))-9=-4#

#:.5-9=-4#

#:.color(blue)(-4=-4#

substitute #color(blue)(x=-4/5#

#:.5color(blue)((-4/5))^2+9(color(blue)(-4/5))=-4#

#:.5(color(blue)(-4/5))(color(blue)(-4/5))+9(color(blue)(-4/5))=-4#

#:.5(16/25)-(36/5)=-4#

#:.80/25-36/5=-4/1#

#:.(80-180=-100)/25#

#:.color(blue)(-100=-100#