Question 140640
Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{x^2+4*x+24=0}}} ( notice {{{a=1}}}, {{{b=4}}}, and {{{c=24}}})





{{{x = (-4 +- sqrt( (4)^2-4*1*24 ))/(2*1)}}} Plug in a=1, b=4, and c=24




{{{x = (-4 +- sqrt( 16-4*1*24 ))/(2*1)}}} Square 4 to get 16  




{{{x = (-4 +- sqrt( 16+-96 ))/(2*1)}}} Multiply {{{-4*24*1}}} to get {{{-96}}}




{{{x = (-4 +- sqrt( -80 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-4 +- 4*i*sqrt(5))/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (-4 +- 4*i*sqrt(5))/(2)}}} Multiply 2 and 1 to get 2




After simplifying, the quadratic has roots of


{{{x=-2 + 2*sqrt(5)i}}} or {{{x=-2 - 2*sqrt(5)i}}}