Question 591533


{{{4x^2+x-6=0}}} Start with the given equation.



Notice that the quadratic {{{4x^2+x-6}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=4}}}, {{{B=1}}}, and {{{C=-6}}}



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(4)(-6) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=1}}}, and {{{C=-6}}}



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



{{{x = (-1 +- sqrt( 1--96 ))/(2(4))}}} Multiply {{{4(4)(-6)}}} to get {{{-96}}}



{{{x = (-1 +- sqrt( 1+96 ))/(2(4))}}} Rewrite {{{sqrt(1--96)}}} as {{{sqrt(1+96)}}}



{{{x = (-1 +- sqrt( 97 ))/(2(4))}}} Add {{{1}}} to {{{96}}} to get {{{97}}}



{{{x = (-1 +- sqrt( 97 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (-1+sqrt(97))/(8)}}} or {{{x = (-1-sqrt(97))/(8)}}} Break up the expression.  



So the solutions are {{{x = (-1+sqrt(97))/(8)}}} or {{{x = (-1-sqrt(97))/(8)}}}