Question 143650
I'm assuming that you want to solve for x




{{{x^2+20x=200}}} Start with the given equation



{{{x^2+20x-200=0}}}  Subtract 200 from both sides. 


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+20*x-200=0}}} ( notice {{{a=1}}}, {{{b=20}}}, and {{{c=-200}}})





{{{x = (-20 +- sqrt( (20)^2-4*1*-200 ))/(2*1)}}} Plug in a=1, b=20, and c=-200




{{{x = (-20 +- sqrt( 400-4*1*-200 ))/(2*1)}}} Square 20 to get 400  




{{{x = (-20 +- sqrt( 400+800 ))/(2*1)}}} Multiply {{{-4*-200*1}}} to get {{{800}}}




{{{x = (-20 +- sqrt( 1200 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-20 +- 20*sqrt(3))/(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 = (-20 +- 20*sqrt(3))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-20 + 20*sqrt(3))/2}}} or {{{x = (-20 - 20*sqrt(3))/2}}}



Now break up the fraction



{{{x=-20/2+20*sqrt(3)/2}}} or {{{x=-20/2-20*sqrt(3)/2}}}



Simplify



{{{x=-10+10*sqrt(3)}}} or {{{x=-10-10*sqrt(3)}}}




So our answers are



{{{x=-10+10*sqrt(3)}}} or {{{x=-10-10*sqrt(3)}}}



which approximate to 



{{{x=7.32050808}}} or {{{x=-27.32050808}}}