Question 201632


{{{x^2+x-42=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+x-42}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=1}}}, and {{{C=-42}}}



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



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



{{{x = (-(1) +- sqrt( (1)^2-4(1)(-42) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=1}}}, and {{{C=-42}}}



{{{x = (-1 +- sqrt( 1-4(1)(-42) ))/(2(1))}}} Square {{{1}}} to get {{{1}}}. 



{{{x = (-1 +- sqrt( 1--168 ))/(2(1))}}} Multiply {{{4(1)(-42)}}} to get {{{-168}}}



{{{x = (-1 +- sqrt( 1+168 ))/(2(1))}}} Rewrite {{{sqrt(1--168)}}} as {{{sqrt(1+168)}}}



{{{x = (-1 +- sqrt( 169 ))/(2(1))}}} Add {{{1}}} to {{{168}}} to get {{{169}}}



{{{x = (-1 +- sqrt( 169 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-1 +- 13)/(2)}}} Take the square root of {{{169}}} to get {{{13}}}. 



{{{x = (-1 + 13)/(2)}}} or {{{x = (-1 - 13)/(2)}}} Break up the expression. 



{{{x = (12)/(2)}}} or {{{x =  (-14)/(2)}}} Combine like terms. 



{{{x = 6}}} or {{{x = -7}}} Simplify. 



So the solutions are {{{x = 6}}} or {{{x = -7}}}