Question 186440
{{{2x^2+6x+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=6}}}, 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 = (-(6) +- sqrt( (6)^2-4(2)(5) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=6}}}, and {{{c=5}}}



{{{x = (-6 +- sqrt( 36 - 4(2)(5) ))/(2(2))}}} Square {{{6}}} to get {{{36}}}. 



{{{x = (-6 +- sqrt( 36 - 40))/(2(2))}}} Multiply {{{4(2)(5)}}} to get {{{40}}}



{{{x = (-6 +- sqrt(-4))/(2(2))}}} Subtract {{{40}}} from {{{36}}} to get {{{-4}}}



{{{x = (-6 +- sqrt( -4 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-6 +- 2*i)/(4)}}} Take the square root of {{{-4}}} to get {{{2*i}}}. 



{{{x = (-6 + 2*i)/(4)}}} or {{{x = (-6 - 2*i)/(4)}}} Break up the expression. 



{{{x = (-6)/(4) + (2*i)/(4)}}} or {{{x =  (-6)/(4) - (2*i)/(4)}}} Break up the fraction for each case. 



{{{x = -3/2+(1/2)*i}}} or {{{x =  -3/2-(1/2)*i}}} Reduce. 



So the answers are {{{x = -3/2+(1/2)*i}}} or {{{x =  -3/2-(1/2)*i}}}