Question 179010


{{{5x^2+6x+12=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=5}}}, {{{b=6}}}, and {{{c=12}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(6) +- sqrt( (6)^2-4(5)(12) ))/(2(5))}}} Plug in  {{{a=5}}}, {{{b=6}}}, and {{{c=12}}}



{{{x = (-6 +- sqrt( 36-4(5)(12) ))/(2(5))}}} Square {{{6}}} to get {{{36}}}. 



{{{x = (-6 +- sqrt( 36-240 ))/(2(5))}}} Multiply {{{4(5)(12)}}} to get {{{240}}}



{{{x = (-6 +- sqrt( -204 ))/(2(5))}}} Subtract {{{240}}} from {{{36}}} to get {{{-204}}}



{{{x = (-6 +- sqrt( -204 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{x = (-6 +- 2i*sqrt(51))/(10)}}} 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 = (-6+2i*sqrt(51))/(10)}}} or {{{x = (-6-2i*sqrt(51))/(10)}}} Break up the expression.  



{{{x = (-3+i*sqrt(51))/(5)}}} or {{{x = (-3-i*sqrt(51))/(5)}}} Reduce.



So the answers are {{{x = (-3+i*sqrt(51))/(5)}}} or {{{x = (-3-i*sqrt(51))/(5)}}}