Question 194983


{{{6x^2+38x+12=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=6}}}, {{{b=38}}}, and {{{c=12}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(38) +- sqrt( (38)^2-4(6)(12) ))/(2(6))}}} Plug in  {{{a=6}}}, {{{b=38}}}, and {{{c=12}}}



{{{x = (-38 +- sqrt( 1444-4(6)(12) ))/(2(6))}}} Square {{{38}}} to get {{{1444}}}. 



{{{x = (-38 +- sqrt( 1444-288 ))/(2(6))}}} Multiply {{{4(6)(12)}}} to get {{{288}}}



{{{x = (-38 +- sqrt( 1156 ))/(2(6))}}} Subtract {{{288}}} from {{{1444}}} to get {{{1156}}}



{{{x = (-38 +- sqrt( 1156 ))/(12)}}} Multiply {{{2}}} and {{{6}}} to get {{{12}}}. 



{{{x = (-38 +- 34)/(12)}}} Take the square root of {{{1156}}} to get {{{34}}}. 



{{{x = (-38 + 34)/(12)}}} or {{{x = (-38 - 34)/(12)}}} Break up the expression. 



{{{x = (-4)/(12)}}} or {{{x =  (-72)/(12)}}} Combine like terms. 



{{{x = -1/3}}} or {{{x = -6}}} Simplify. 



So the answers are {{{x = -1/3}}} or {{{x = -6}}}