Question 116981
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+8*x+25=0}}} ( notice {{{a=1}}}, {{{b=8}}}, and {{{c=25}}})





{{{x = (-8 +- sqrt( (8)^2-4*1*25 ))/(2*1)}}} Plug in a=1, b=8, and c=25




{{{x = (-8 +- sqrt( 64-4*1*25 ))/(2*1)}}} Square 8 to get 64  




{{{x = (-8 +- sqrt( 64+-100 ))/(2*1)}}} Multiply {{{-4*25*1}}} to get {{{-100}}}




{{{x = (-8 +- sqrt( -36 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-8 +- 6*i)/(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 = (-8 +- 6*i)/(2)}}} Multiply 2 and 1 to get 2




After simplifying, the quadratic has roots of


{{{x=-4 + 3*i}}} or {{{x=-4 - 3*i}}}



So the answer is D)