Question 448095

{{{2x^2+4x+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=4}}}, 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 = (-(4) +- sqrt( (4)^2-4(2)(1) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=4}}}, and {{{c=1}}}



{{{x = (-4 +- sqrt( 16-4(2)(1) ))/(2(2))}}} Square {{{4}}} to get {{{16}}}. 



{{{x = (-4 +- sqrt( 16-8 ))/(2(2))}}} Multiply {{{4(2)(1)}}} to get {{{8}}}



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



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



{{{x = (-4 +- 2*sqrt(2))/(4)}}} Simplify the square root  



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



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



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