Question 214878


{{{x^2+10x+1=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+10x+1}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=10}}}, 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 = (-(10) +- sqrt( (10)^2-4(1)(1) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=10}}}, and {{{C=1}}}



{{{x = (-10 +- sqrt( 100-4(1)(1) ))/(2(1))}}} Square {{{10}}} to get {{{100}}}. 



{{{x = (-10 +- sqrt( 100-4 ))/(2(1))}}} Multiply {{{4(1)(1)}}} to get {{{4}}}



{{{x = (-10 +- sqrt( 96 ))/(2(1))}}} Subtract {{{4}}} from {{{100}}} to get {{{96}}}



{{{x = (-10 +- sqrt( 96 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-10 +- 4*sqrt(6))/(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 = (-10)/(2) +- (4*sqrt(6))/(2)}}} Break up the fraction.  



{{{x = -5 +- 2*sqrt(6)}}} Reduce.  



{{{x = -5+2*sqrt(6)}}} or {{{x = -5-2*sqrt(6)}}} Break up the expression.  



So the solutions are {{{x = -5+2*sqrt(6)}}} or {{{x = -5-2*sqrt(6)}}} 



which approximate to {{{x=-0.101}}} or {{{x=-9.899}}}