Question 187428


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



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=4}}}, {{{b=2}}}, and {{{c=8}}}



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(4)(8) ))/(2(4))}}} Plug in  {{{a=4}}}, {{{b=2}}}, and {{{c=8}}}



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



{{{x = (-2 +- sqrt( 4-128 ))/(2(4))}}} Multiply {{{4(4)(8)}}} to get {{{128}}}



{{{x = (-2 +- sqrt( -124 ))/(2(4))}}} Subtract {{{128}}} from {{{4}}} to get {{{-124}}}



{{{x = (-2 +- sqrt( -124 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (-2 +- 2i*sqrt(31))/(8)}}} 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+2i*sqrt(31))/(8)}}} or {{{x = (-2-2i*sqrt(31))/(8)}}} Break up the expression.  



{{{x = (-1+i*sqrt(31))/(4)}}} or {{{x = (-1-i*sqrt(31))/(4)}}} Reduce.




So the answers are {{{x = (-1+i*sqrt(31))/(4)}}} or {{{x = (-1-i*sqrt(31))/(4)}}}