Question 200025


{{{x^2+14x-31=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+14x-31}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=14}}}, and {{{C=-31}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(14) +- sqrt( (14)^2-4(1)(-31) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=14}}}, and {{{C=-31}}}



{{{x = (-14 +- sqrt( 196-4(1)(-31) ))/(2(1))}}} Square {{{14}}} to get {{{196}}}. 



{{{x = (-14 +- sqrt( 196--124 ))/(2(1))}}} Multiply {{{4(1)(-31)}}} to get {{{-124}}}



{{{x = (-14 +- sqrt( 196+124 ))/(2(1))}}} Rewrite {{{sqrt(196--124)}}} as {{{sqrt(196+124)}}}



{{{x = (-14 +- sqrt( 320 ))/(2(1))}}} Add {{{196}}} to {{{124}}} to get {{{320}}}



{{{x = (-14 +- sqrt( 320 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-14 +- 8*sqrt(5))/(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 = (-14)/(2) +- (8*sqrt(5))/(2)}}} Break up the fraction.  



{{{x = -7 +- 4*sqrt(5)}}} Reduce.  



{{{x = -7+4*sqrt(5)}}} or {{{x = -7-4*sqrt(5)}}} Break up the expression.  



So the solutions are {{{x = -7+4*sqrt(5)}}} or {{{x = -7-4*sqrt(5)}}} 



which approximate to {{{x=1.944}}} or {{{x=-15.944}}}