Question 574640


{{{-16t^2-12t=2}}} Start with the given equation.



{{{-16t^2-12t-2=0}}} Get every term to the left side.



Notice that the quadratic {{{-16t^2-12t-2}}} is in the form of {{{At^2+Bt+C}}} where {{{A=-16}}}, {{{B=-12}}}, and {{{C=-2}}}



Let's use the quadratic formula to solve for "t":



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



{{{t = (-(-12) +- sqrt( (-12)^2-4(-16)(-2) ))/(2(-16))}}} Plug in  {{{A=-16}}}, {{{B=-12}}}, and {{{C=-2}}}



{{{t = (12 +- sqrt( (-12)^2-4(-16)(-2) ))/(2(-16))}}} Negate {{{-12}}} to get {{{12}}}. 



{{{t = (12 +- sqrt( 144-4(-16)(-2) ))/(2(-16))}}} Square {{{-12}}} to get {{{144}}}. 



{{{t = (12 +- sqrt( 144-128 ))/(2(-16))}}} Multiply {{{4(-16)(-2)}}} to get {{{128}}}



{{{t = (12 +- sqrt( 16 ))/(2(-16))}}} Subtract {{{128}}} from {{{144}}} to get {{{16}}}



{{{t = (12 +- sqrt( 16 ))/(-32)}}} Multiply {{{2}}} and {{{-16}}} to get {{{-32}}}. 



{{{t = (12 +- 4)/(-32)}}} Take the square root of {{{16}}} to get {{{4}}}. 



{{{t = (12 + 4)/(-32)}}} or {{{t = (12 - 4)/(-32)}}} Break up the expression. 



{{{t = (16)/(-32)}}} or {{{t =  (8)/(-32)}}} Combine like terms. 



{{{t = -1/2}}} or {{{t = -1/4}}} Simplify. 



So the solutions are {{{t = -1/2}}} or {{{t = -1/4}}}