Question 188659


{{{x^2+10x+30=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=10}}}, and {{{c=30}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(10) +- sqrt( (10)^2-4(1)(30) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=10}}}, and {{{c=30}}}



{{{x = (-10 +- sqrt( 100-4(1)(30) ))/(2(1))}}} Square {{{10}}} to get {{{100}}}. 



{{{x = (-10 +- sqrt( 100-120 ))/(2(1))}}} Multiply {{{4(1)(30)}}} to get {{{120}}}



{{{x = (-10 +- sqrt( -20 ))/(2(1))}}} Subtract {{{120}}} from {{{100}}} to get {{{-20}}}



{{{x = (-10 +- sqrt( -20 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-10 +- 2i*sqrt(5))/(2)}}} 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 = (-10)/(2) +- (2i*sqrt(5))/(2)}}} Break up the fraction.  



{{{x = -5 +- sqrt(5)*i}}} Reduce.  



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



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




So the solutions are in the form {{{a+bi}}} where {{{a=-5}}} (for both solutions), {{{b=sqrt(5)}}} for the first solution, and {{{b=-sqrt(5)}}} for the second solution.