Question 460081


{{{x^2+14x+4=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+14x+4}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=14}}}, 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 = (-(14) +- sqrt( (14)^2-4(1)(4) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=14}}}, and {{{C=4}}}



{{{x = (-14 +- sqrt( 196-4(1)(4) ))/(2(1))}}} Square {{{14}}} to get {{{196}}}. 



{{{x = (-14 +- sqrt( 196-16 ))/(2(1))}}} Multiply {{{4(1)(4)}}} to get {{{16}}}



{{{x = (-14 +- sqrt( 180 ))/(2(1))}}} Subtract {{{16}}} from {{{196}}} to get {{{180}}}



{{{x = (-14 +- sqrt( 180 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-14 +- 6*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 = (-14)/(2) +- (6*sqrt(5))/(2)}}} Break up the fraction.  



{{{x = -7 +- 3*sqrt(5)}}} Reduce.  



{{{x = -7+3*sqrt(5)}}} or {{{x = -7-3*sqrt(5)}}} Break up the expression.  



So the solutions are {{{x = -7+3*sqrt(5)}}} or {{{x = -7-3*sqrt(5)}}}