Question 689013


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



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



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



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



{{{x = (-2 +- sqrt( 4+40 ))/(2(2))}}} Rewrite {{{sqrt(4--40)}}} as {{{sqrt(4+40)}}}



{{{x = (-2 +- sqrt( 44 ))/(2(2))}}} Add {{{4}}} to {{{40}}} to get {{{44}}}



{{{x = (-2 +- sqrt( 44 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-2 +- 2*sqrt(11))/(4)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (-2+2*sqrt(11))/(4)}}} or {{{x = (-2-2*sqrt(11))/(4)}}} Break up the expression.  



{{{x = (-1+sqrt(11))/(2)}}} or {{{x = (-1-sqrt(11))/(2)}}} Reduce



So the solutions are {{{x = (-1+sqrt(11))/(2)}}} or {{{x = (-1-sqrt(11))/(2)}}}