How do you solve #2x^2+3x+1=0#?

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Answer 1 · 2016-05-16T17:05:10

The solutions are:
#x = - 1/2#

#x = - 1 #

#2x^2 + 3x + 1 = 0 #
4
The equation is of the form #color(blue)(ax^2+bx+c=0# where:

#a=2, b=3, c=1#

The Discriminant is given by:

#Delta=b^2-4*a*c#

# = (3)^2-(4 * 2 * 1)#

# = 9 - 8 = 1 #

The solutions are found using the formula
#x=(-b+-sqrtDelta)/(2*a)#

#x = ((-3)+-sqrt(1))/(2*2) = (-3 +- ( 1) )/4#

#x = (- 3 + 1 ) / 4 = -2/ 4 = - 1/2#

#x = (- 3 - 1 ) / 4 = -4/4 = - 1 #