Question 134353
Let's use the quadratic formula to solve for t:



Starting with the general quadratic


{{{at^2+bt+c=0}}}


the general solution using the quadratic equation is:


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




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





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




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




{{{t = (-2 +- sqrt( 4+-8 ))/(2*1)}}} Multiply {{{-4*2*1}}} to get {{{-8}}}




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




{{{t = (-2 +- 2*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>)




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



After simplifying, the quadratic has roots of


{{{t=-1 + i}}} or {{{t=-1 - i}}}