How do you solve #2x^2+5x+2=0#?

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Answer 1 · 2015-07-12T16:36:11

#color(blue)(x=-1/2,x=-2#

#2x^2+5x+2#

Here, we can first factorise the expression and then find the solutions.

Factorising by splitting middle term

#2x^2+color(green)(5x)+2=0#

#2x^2+color(green)(4x +x)+2=0#

#2x(x+2)+1(x+2)=0#

#(2x+1)(x+2)=0#
Now, we equate the factors with zero to find the solutions.

#2x+1=0, color(blue)(x=-1/2#
#x+2=0 , color(blue)(x=-2#