Question 280212


{{{x^2+6x-1=0}}} Start with the given equation.



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



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



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



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



{{{x = (-6 +- sqrt( 36+4 ))/(2(1))}}} Rewrite {{{sqrt(36--4)}}} as {{{sqrt(36+4)}}}



{{{x = (-6 +- sqrt( 40 ))/(2(1))}}} Add {{{36}}} to {{{4}}} to get {{{40}}}



{{{x = (-6 +- sqrt( 40 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-6 +- 2*sqrt(10))/(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(10))/(2)}}} Break up the fraction.  



{{{x = -3 +- sqrt(10)}}} Reduce.  



{{{x = -3+sqrt(10)}}} or {{{x = -3-sqrt(10)}}} Break up the expression.  



So the solutions are {{{x = -3+sqrt(10)}}} or {{{x = -3-sqrt(10)}}}