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

2 Answers
ali ergin
Oct 4, 2016

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

Explanation:

"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}

Veddesh phi
Oct 4, 2016

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

Explanation:

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

Solve using Quadratic formula

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)

Related questions
See all questions in Quadratic Formula
Impact of this question
20726 views around the world
You can reuse this answer
Creative Commons License