Question 613860


{{{x^2-3x-28=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-3x-28}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-3}}}, and {{{C=-28}}}



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



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



{{{x = (-(-3) +- sqrt( (-3)^2-4(1)(-28) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-3}}}, and {{{C=-28}}}



{{{x = (3 +- sqrt( (-3)^2-4(1)(-28) ))/(2(1))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{x = (3 +- sqrt( 9-4(1)(-28) ))/(2(1))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9--112 ))/(2(1))}}} Multiply {{{4(1)(-28)}}} to get {{{-112}}}



{{{x = (3 +- sqrt( 9+112 ))/(2(1))}}} Rewrite {{{sqrt(9--112)}}} as {{{sqrt(9+112)}}}



{{{x = (3 +- sqrt( 121 ))/(2(1))}}} Add {{{9}}} to {{{112}}} to get {{{121}}}



{{{x = (3 +- sqrt( 121 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (3 +- 11)/(2)}}} Take the square root of {{{121}}} to get {{{11}}}. 



{{{x = (3 + 11)/(2)}}} or {{{x = (3 - 11)/(2)}}} Break up the expression. 



{{{x = (14)/(2)}}} or {{{x =  (-8)/(2)}}} Combine like terms. 



{{{x = 7}}} or {{{x = -4}}} Simplify. 



So the solutions are {{{x = 7}}} or {{{x = -4}}} 

  


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