Question 309947


{{{2y^2-3y-6=0}}} Start with the given equation.



Notice that the quadratic {{{2y^2-3y-6}}} is in the form of {{{Ay^2+By+C}}} where {{{A=2}}}, {{{B=-3}}}, and {{{C=-6}}}



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



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



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



{{{y = (3 +- sqrt( (-3)^2-4(2)(-6) ))/(2(2))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{y = (3 +- sqrt( 9-4(2)(-6) ))/(2(2))}}} Square {{{-3}}} to get {{{9}}}. 



{{{y = (3 +- sqrt( 9--48 ))/(2(2))}}} Multiply {{{4(2)(-6)}}} to get {{{-48}}}



{{{y = (3 +- sqrt( 9+48 ))/(2(2))}}} Rewrite {{{sqrt(9--48)}}} as {{{sqrt(9+48)}}}



{{{y = (3 +- sqrt( 57 ))/(2(2))}}} Add {{{9}}} to {{{48}}} to get {{{57}}}



{{{y = (3 +- sqrt( 57 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{y = (3+sqrt(57))/(4)}}} or {{{y = (3-sqrt(57))/(4)}}} Break up the expression.  



So the solutions are {{{y = (3+sqrt(57))/(4)}}} or {{{y = (3-sqrt(57))/(4)}}}