Question 171866


{{{3z^2+13z+12=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{az^2+bz+c}}} where {{{a=3}}}, {{{b=13}}}, and {{{c=12}}}



Let's use the quadratic formula to solve for z



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



{{{z = (-(13) +- sqrt( (13)^2-4(3)(12) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=13}}}, and {{{c=12}}}



{{{z = (-13 +- sqrt( 169-4(3)(12) ))/(2(3))}}} Square {{{13}}} to get {{{169}}}. 



{{{z = (-13 +- sqrt( 169-144 ))/(2(3))}}} Multiply {{{4(3)(12)}}} to get {{{144}}}



{{{z = (-13 +- sqrt( 25 ))/(2(3))}}} Subtract {{{144}}} from {{{169}}} to get {{{25}}}



{{{z = (-13 +- sqrt( 25 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{z = (-13 +- 5)/(6)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{z = (-13 + 5)/(6)}}} or {{{z = (-13 - 5)/(6)}}} Break up the expression. 



{{{z = (-8)/(6)}}} or {{{z =  (-18)/(6)}}} Combine like terms. 



{{{z = -4/3}}} or {{{z = -3}}} Simplify. 



So the answers are {{{z = -4/3}}} or {{{z = -3}}}