Question 580522


{{{2x^2+9x-5=0}}} Start with the given equation.



Notice that the quadratic {{{2x^2+9x-5}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=9}}}, 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 = (-(9) +- sqrt( (9)^2-4(2)(-5) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=9}}}, and {{{C=-5}}}



{{{x = (-9 +- sqrt( 81-4(2)(-5) ))/(2(2))}}} Square {{{9}}} to get {{{81}}}. 



{{{x = (-9 +- sqrt( 81--40 ))/(2(2))}}} Multiply {{{4(2)(-5)}}} to get {{{-40}}}



{{{x = (-9 +- sqrt( 81+40 ))/(2(2))}}} Rewrite {{{sqrt(81--40)}}} as {{{sqrt(81+40)}}}



{{{x = (-9 +- sqrt( 121 ))/(2(2))}}} Add {{{81}}} to {{{40}}} to get {{{121}}}



{{{x = (-9 +- sqrt( 121 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-9 +- 11)/(4)}}} Take the square root of {{{121}}} to get {{{11}}}. 



{{{x = (-9 + 11)/(4)}}} or {{{x = (-9 - 11)/(4)}}} Break up the expression. 



{{{x = (2)/(4)}}} or {{{x =  (-20)/(4)}}} Combine like terms. 



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



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