Question 417121


{{{2x^2-3x-5=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=-3}}}, and {{{c=-5}}}



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(2)(-5) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=-3}}}, and {{{c=-5}}}



{{{x = (3 +- sqrt( (-3)^2-4(2)(-5) ))/(2(2))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{x = (3 +- sqrt( 9-4(2)(-5) ))/(2(2))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9--40 ))/(2(2))}}} Multiply {{{4(2)(-5)}}} to get {{{-40}}}



{{{x = (3 +- sqrt( 9+40 ))/(2(2))}}} Rewrite {{{sqrt(9--40)}}} as {{{sqrt(9+40)}}}



{{{x = (3 +- sqrt( 49 ))/(2(2))}}} Add {{{9}}} to {{{40}}} to get {{{49}}}



{{{x = (3 +- sqrt( 49 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (3 +- 7)/(4)}}} Take the square root of {{{49}}} to get {{{7}}}. 



{{{x = (3 + 7)/(4)}}} or {{{x = (3 - 7)/(4)}}} Break up the expression. 



{{{x = (10)/(4)}}} or {{{x =  (-4)/(4)}}} Combine like terms. 



{{{x = 5/2}}} or {{{x = -1}}} Simplify. 



So the answers are {{{x = 5/2}}} or {{{x = -1}}} 

  


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