Question 635895


{{{3x^2+8x+4=0}}} Start with the given equation.



Notice that the quadratic {{{3x^2+8x+4}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=8}}}, and {{{C=4}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(8) +- sqrt( (8)^2-4(3)(4) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=8}}}, and {{{C=4}}}



{{{x = (-8 +- sqrt( 64-4(3)(4) ))/(2(3))}}} Square {{{8}}} to get {{{64}}}. 



{{{x = (-8 +- sqrt( 64-48 ))/(2(3))}}} Multiply {{{4(3)(4)}}} to get {{{48}}}



{{{x = (-8 +- sqrt( 16 ))/(2(3))}}} Subtract {{{48}}} from {{{64}}} to get {{{16}}}



{{{x = (-8 +- sqrt( 16 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (-8 +- 4)/(6)}}} Take the square root of {{{16}}} to get {{{4}}}. 



{{{x = (-8 + 4)/(6)}}} or {{{x = (-8 - 4)/(6)}}} Break up the expression. 



{{{x = (-4)/(6)}}} or {{{x =  (-12)/(6)}}} Combine like terms. 



{{{x = -2/3}}} or {{{x = -2}}} Simplify. 



So the solutions are {{{x = -2/3}}} or {{{x = -2}}}