Question 177509


{{{2x^2+5x-1=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=5}}}, 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 = (-(5) +- sqrt( (5)^2-4(2)(-1) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=5}}}, and {{{c=-1}}}



{{{x = (-5 +- sqrt( 25-4(2)(-1) ))/(2(2))}}} Square {{{5}}} to get {{{25}}}. 



{{{x = (-5 +- sqrt( 25--8 ))/(2(2))}}} Multiply {{{4(2)(-1)}}} to get {{{-8}}}



{{{x = (-5 +- sqrt( 25+8 ))/(2(2))}}} Rewrite {{{sqrt(25--8)}}} as {{{sqrt(25+8)}}}



{{{x = (-5 +- sqrt( 33 ))/(2(2))}}} Add {{{25}}} to {{{8}}} to get {{{33}}}



{{{x = (-5 +- sqrt( 33 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-5+sqrt(33))/(4)}}} or {{{x = (-5-sqrt(33))/(4)}}} Break up the expression.  



So the answers are {{{x = (-5+sqrt(33))/(4)}}} or {{{x = (-5-sqrt(33))/(4)}}} 



which approximate to {{{x=0.186}}} or {{{x=-2.686}}}