Question 605414


{{{16d^2+36d+18=0}}} Start with the given equation.



Notice that the quadratic {{{16d^2+36d+18}}} is in the form of {{{Ad^2+Bd+C}}} where {{{A=16}}}, {{{B=36}}}, and {{{C=18}}}



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



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



{{{d = (-(36) +- sqrt( (36)^2-4(16)(18) ))/(2(16))}}} Plug in  {{{A=16}}}, {{{B=36}}}, and {{{C=18}}}



{{{d = (-36 +- sqrt( 1296-4(16)(18) ))/(2(16))}}} Square {{{36}}} to get {{{1296}}}. 



{{{d = (-36 +- sqrt( 1296-1152 ))/(2(16))}}} Multiply {{{4(16)(18)}}} to get {{{1152}}}



{{{d = (-36 +- sqrt( 144 ))/(2(16))}}} Subtract {{{1152}}} from {{{1296}}} to get {{{144}}}



{{{d = (-36 +- sqrt( 144 ))/(32)}}} Multiply {{{2}}} and {{{16}}} to get {{{32}}}. 



{{{d = (-36 +- 12)/(32)}}} Take the square root of {{{144}}} to get {{{12}}}. 



{{{d = (-36 + 12)/(32)}}} or {{{d = (-36 - 12)/(32)}}} Break up the expression. 



{{{d = (-24)/(32)}}} or {{{d =  (-48)/(32)}}} Combine like terms. 



{{{d = -3/4}}} or {{{d = -3/2}}} Simplify. 



So the solutions are {{{d = -3/4}}} or {{{d = -3/2}}} 

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