How do you solve # 3x^2 + 5x + 2 = 0#?

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Answer 1 · 2016-05-29T16:16:39

The solutions are:

#color(blue)(x = -2/3#

#color(blue)(x = -1#

#3x^2 +5x + 2 = 0 #

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

#a=3, b=5, c=2#

The Discriminant is given by:

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

# = (5)^2-(4 * 3 * 2)#

# = 25 - 24 = 1#

The solutions are found using the formula:

#x=(-b+-sqrtDelta)/(2*a)#

#x = ((-5)+-sqrt(1))/(2*3) = (-5 +- 1 )/8#

#x = ( -5 + 1 ) / 6 = -4/6 = -2/3#

#x = ( -5 - 1 ) / 6 = -6/6 = -1#