Question 266175


{{{x^2+12x+4=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+12x+4}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=12}}}, 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 = (-(12) +- sqrt( (12)^2-4(1)(4) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=12}}}, and {{{C=4}}}



{{{x = (-12 +- sqrt( 144-4(1)(4) ))/(2(1))}}} Square {{{12}}} to get {{{144}}}. 



{{{x = (-12 +- sqrt( 144-16 ))/(2(1))}}} Multiply {{{4(1)(4)}}} to get {{{16}}}



{{{x = (-12 +- sqrt( 128 ))/(2(1))}}} Subtract {{{16}}} from {{{144}}} to get {{{128}}}



{{{x = (-12 +- sqrt( 128 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-12 +- 8*sqrt(2))/(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 = (-12)/(2) +- (8*sqrt(2))/(2)}}} Break up the fraction.  



{{{x = -6 +- 4*sqrt(2)}}} Reduce.  



{{{x = -6+4*sqrt(2)}}} or {{{x = -6-4*sqrt(2)}}} Break up the expression.  



So the solutions are {{{x = -6+4*sqrt(2)}}} or {{{x = -6-4*sqrt(2)}}}