Question 600338


{{{4x^2+8x+5=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=4}}}, {{{b=8}}}, 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 = (-(8) +- sqrt( (8)^2-4(4)(5) ))/(2(4))}}} Plug in  {{{a=4}}}, {{{b=8}}}, and {{{c=5}}}



{{{x = (-8 +- sqrt( 64-4(4)(5) ))/(2(4))}}} Square {{{8}}} to get {{{64}}}. 



{{{x = (-8 +- sqrt( 64-80 ))/(2(4))}}} Multiply {{{4(4)(5)}}} to get {{{80}}}



{{{x = (-8 +- sqrt( -16 ))/(2(4))}}} Subtract {{{80}}} from {{{64}}} to get {{{-16}}}



{{{x = (-8 +- sqrt( -16 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (-8 +- 4*i)/(8)}}} Take the square root of {{{-16}}} to get {{{4*i}}}. 



{{{x = (-8 + 4*i)/(8)}}} or {{{x = (-8 - 4*i)/(8)}}} Break up the expression. 



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



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