```Question 266175

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 = (-(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)}}} ```