Question 481217


{{{2x^2+3x-1=0}}} Start with the given equation.



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



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)(-1) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=3}}}, and {{{C=-1}}}



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



{{{x = (-3 +- sqrt( 9--8 ))/(2(2))}}} Multiply {{{4(2)(-1)}}} to get {{{-8}}}



{{{x = (-3 +- sqrt( 9+8 ))/(2(2))}}} Rewrite {{{sqrt(9--8)}}} as {{{sqrt(9+8)}}}



{{{x = (-3 +- sqrt( 17 ))/(2(2))}}} Add {{{9}}} to {{{8}}} to get {{{17}}}



{{{x = (-3 +- sqrt( 17 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-3+sqrt(17))/(4)}}} or {{{x = (-3-sqrt(17))/(4)}}} Break up the expression.  



So the solutions are {{{x = (-3+sqrt(17))/(4)}}} or {{{x = (-3-sqrt(17))/(4)}}}