How do you solve #8x ^ { 2} + 40x = - 50#?

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Answer 1 · 2017-12-02T10:15:04

#x=-2.5#

#8x^2+40x=-50#

#:.8x^2+40x+50=0#

#:.2(4x^2+20x+25)=0#

#:.2(2x+5)(2x+5)=0#

#:.2x+5=0#

#:.2x=-5#

#:.x=-5/2#

#:.x=-2.5#

~~~~~~~~~~~~

check:-

substitute #x=-2.5#

#:.8(-2.5)^2+40(-2.5)=-50#

#:.8(6.25)-100=-50#

#:.50-100=-50#

#:.-50=-50#