Question 596063


{{{12c^2+8c+1=0}}} Start with the given equation.



Notice that the quadratic {{{12c^2+8c+1}}} is in the form of {{{Ac^2+Bc+C}}} where {{{A=12}}}, {{{B=8}}}, and {{{C=1}}}



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



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



{{{c = (-(8) +- sqrt( (8)^2-4(12)(1) ))/(2(12))}}} Plug in  {{{A=12}}}, {{{B=8}}}, and {{{C=1}}}



{{{c = (-8 +- sqrt( 64-4(12)(1) ))/(2(12))}}} Square {{{8}}} to get {{{64}}}. 



{{{c = (-8 +- sqrt( 64-48 ))/(2(12))}}} Multiply {{{4(12)(1)}}} to get {{{48}}}



{{{c = (-8 +- sqrt( 16 ))/(2(12))}}} Subtract {{{48}}} from {{{64}}} to get {{{16}}}



{{{c = (-8 +- sqrt( 16 ))/(24)}}} Multiply {{{2}}} and {{{12}}} to get {{{24}}}. 



{{{c = (-8 +- 4)/(24)}}} Take the square root of {{{16}}} to get {{{4}}}. 



{{{c = (-8 + 4)/(24)}}} or {{{c = (-8 - 4)/(24)}}} Break up the expression. 



{{{c = (-4)/(24)}}} or {{{c =  (-12)/(24)}}} Combine like terms. 



{{{c = -1/6}}} or {{{c = -1/2}}} Simplify. 



So the solutions are {{{c = -1/6}}} or {{{c = -1/2}}}