How do you solve #3x^2 + 8x + 4 = 0# using the quadratic formula?

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Answer 1 · 2016-10-04T13:12:40

#x_"1,2"={-2/3,-2}#

#"given : "3x^2+8x+4=0#

#ax^2+bx+c=0#

#a=3#
#b=8#
#c=4#

#Delta=sqrt(b^2-4*a*c)#

#Delta=sqrt(8^2-4*3*4)#

#Delta=sqrt(64-48)#

#Delta=sqrt16#

#x_"1"=(-b-Delta)/(2*a)" ; "x_1=(-8-4)/(2*3)" ; "x_1=-12/6=-2#

#x_2=(-b+Delta)/(2*a)" ; "x_2=(-8+4)/(2*3)" ; "x_2=-4/6=-2/3#

#x_"1,2"={-2/3,-2}#

Answer 2 · 2016-10-04T15:02:29

#color(green)(x=(-2,-2/3)#

#color(blue)(3x^2+8x+4=0#

#color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)#

#"a,b and c are the coefficients of the terms"#

Assign the values

#color(orange)(a=3#

#color(orange)(b=8#

#color(orange)(c=4#

Solve

#:.x=(-8+-sqrt(8^2-4(3)(4)))/(2(3))#

#rarr(-8+-sqrt(64-48))/(6)#

#rarr(-8+-sqrt(16))/(6)#

#rarr(-8+-sqrt(4*4))/(6)#

#rarr(-8+-4)/(6)#

We can divide the equation into two

#color(violet)((-8-4)/6=-12/6=-2#

#color(purple)((-8+4)/6=-4/6=-2/3#

#:.x=(-2,-2/3)#