```Question 133182
Let's use the quadratic formula to solve for x:

{{{ax^2+bx+c=0}}}

the general solution using the quadratic equation is:

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}

So lets solve {{{x^2+6*x-4=0}}} ( notice {{{a=1}}}, {{{b=6}}}, and {{{c=-4}}})

{{{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*-4*1}}} to get {{{16}}}

{{{x = (-6 +- sqrt( 52 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)

{{{x = (-6 +- 2*sqrt(13))/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)

{{{x = (-6 +- 2*sqrt(13))/2}}} Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

{{{x = (-6 + 2*sqrt(13))/2}}} or {{{x = (-6 - 2*sqrt(13))/2}}}

Now break up the fraction

{{{x=-6/2+2*sqrt(13)/2}}} or {{{x=-6/2-2*sqrt(13)/2}}}

Simplify

{{{x=-3+sqrt(13)}}} or {{{x=-3-sqrt(13)}}}

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