Question 613730


{{{x^2+6x+4=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+6x+4}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=6}}}, and {{{C=4}}}



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(1)(4) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=6}}}, and {{{C=4}}}



{{{x = (-6 +- sqrt( 36-4(1)(4) ))/(2(1))}}} Square {{{6}}} to get {{{36}}}. 



{{{x = (-6 +- sqrt( 36-16 ))/(2(1))}}} Multiply {{{4(1)(4)}}} to get {{{16}}}



{{{x = (-6 +- sqrt( 20 ))/(2(1))}}} Subtract {{{16}}} from {{{36}}} to get {{{20}}}



{{{x = (-6 +- sqrt( 20 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-6 +- 2*sqrt(5))/(2)}}} 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 = (-6)/(2) +- (2*sqrt(5))/(2)}}} Break up the fraction.  



{{{x = -3 +- sqrt(5)}}} Reduce.  



{{{x = -3+sqrt(5)}}} or {{{x = -3-sqrt(5)}}} Break up the expression.  



So the exact solutions are {{{x = -3+sqrt(5)}}} or {{{x = -3-sqrt(5)}}} 



Then use a calculator to find the approximate solutions to be {{{x=-0.7639320225}}} or {{{x=-5.236067978}}}



Then round those approximate solutions to three places to get {{{x=-0.764}}} or {{{x=-5.236}}}

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