Question 360341


{{{4x^2+5x+6=0}}} Start with the given equation.



Notice that the quadratic {{{4x^2+5x+6}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=4}}}, {{{B=5}}}, 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 = (-(5) +- sqrt( (5)^2-4(4)(6) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=5}}}, and {{{C=6}}}



{{{x = (-5 +- sqrt( 25-4(4)(6) ))/(2(4))}}} Square {{{5}}} to get {{{25}}}. 



{{{x = (-5 +- sqrt( 25-96 ))/(2(4))}}} Multiply {{{4(4)(6)}}} to get {{{96}}}



{{{x = (-5 +- sqrt( -71 ))/(2(4))}}} Subtract {{{96}}} from {{{25}}} to get {{{-71}}}



{{{x = (-5 +- sqrt( -71 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (-5 +- i*sqrt(71))/(8)}}} 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 = (-5+i*sqrt(71))/(8)}}} or {{{x = (-5-i*sqrt(71))/(8)}}} Break up the expression.  



So the solutions are {{{x = (-5+i*sqrt(71))/(8)}}} or {{{x = (-5-i*sqrt(71))/(8)}}}