Question 206893


{{{-6z^2+7z+3=0}}} Start with the given equation.



Notice that the quadratic {{{-6z^2+7z+3}}} is in the form of {{{Az^2+Bz+C}}} where {{{A=-6}}}, {{{B=7}}}, and {{{C=3}}}



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



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



{{{z = (-(7) +- sqrt( (7)^2-4(-6)(3) ))/(2(-6))}}} Plug in  {{{A=-6}}}, {{{B=7}}}, and {{{C=3}}}



{{{z = (-7 +- sqrt( 49-4(-6)(3) ))/(2(-6))}}} Square {{{7}}} to get {{{49}}}. 



{{{z = (-7 +- sqrt( 49--72 ))/(2(-6))}}} Multiply {{{4(-6)(3)}}} to get {{{-72}}}



{{{z = (-7 +- sqrt( 49+72 ))/(2(-6))}}} Rewrite {{{sqrt(49--72)}}} as {{{sqrt(49+72)}}}



{{{z = (-7 +- sqrt( 121 ))/(2(-6))}}} Add {{{49}}} to {{{72}}} to get {{{121}}}



{{{z = (-7 +- sqrt( 121 ))/(-12)}}} Multiply {{{2}}} and {{{-6}}} to get {{{-12}}}. 



{{{z = (-7 +- 11)/(-12)}}} Take the square root of {{{121}}} to get {{{11}}}. 



{{{z = (-7 + 11)/(-12)}}} or {{{z = (-7 - 11)/(-12)}}} Break up the expression. 



{{{z = (4)/(-12)}}} or {{{z =  (-18)/(-12)}}} Combine like terms. 



{{{z = -1/3}}} or {{{z = 3/2}}} Simplify. 



So the solutions are {{{z = -1/3}}} or {{{z = 3/2}}}