Question 506795


{{{2x^2+5x=12}}} Start with the given equation.



{{{2x^2+5x-12=0}}} Get every term to the left side.



Notice that the quadratic {{{2x^2+5x-12}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=5}}}, and {{{C=-12}}}



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



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



{{{x = (-(5) +- sqrt( (5)^2-4(2)(-12) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=5}}}, and {{{C=-12}}}



{{{x = (-5 +- sqrt( 25-4(2)(-12) ))/(2(2))}}} Square {{{5}}} to get {{{25}}}. 



{{{x = (-5 +- sqrt( 25--96 ))/(2(2))}}} Multiply {{{4(2)(-12)}}} to get {{{-96}}}



{{{x = (-5 +- sqrt( 25+96 ))/(2(2))}}} Rewrite {{{sqrt(25--96)}}} as {{{sqrt(25+96)}}}



{{{x = (-5 +- sqrt( 121 ))/(2(2))}}} Add {{{25}}} to {{{96}}} to get {{{121}}}



{{{x = (-5 +- sqrt( 121 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-5 +- 11)/(4)}}} Take the square root of {{{121}}} to get {{{11}}}. 



{{{x = (-5 + 11)/(4)}}} or {{{x = (-5 - 11)/(4)}}} Break up the expression. 



{{{x = (6)/(4)}}} or {{{x =  (-16)/(4)}}} Combine like terms. 



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



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

  
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Thanks,


Jim