Question 294966
{{{(-10+i)/(2+i)}}} Start with the given expression.



{{{((-10+i)/(2+i))((2-i)/(2-i))}}} Multiply the fraction by {{{(2-i)/(2-i)}}}.



{{{((-10+i)(2-i))/((2+i)(2-i))}}} Combine the fractions.



{{{((-10)(2)+(-10)(-i)+(i)(2)+(i)(-i))/((2+i)(2-i))}}} FOIL the numerator.



{{{((-10)(2)+(-10)(-i)+(i)(2)+(i)(-i))/((2)(2)+(2)(-i)+(i)(2)+(i)(-i))}}} FOIL the denominator.



{{{(-20+10i+2i-i^2)/(4-2i+2i-i^2)}}} Multiply.



{{{(-20+10i+2i-(-1))/(4-2i+2i-(-1))}}} Replace {{{i^2}}} with -1.



{{{(-19+12i)/(5)}}} Combine like terms.



{{{(-19)/(5)+((12)/(5))i}}} Break up the fraction.



{{{-19/5+(12/5)i}}} Reduce.



So {{{(-10+i)/(2+i)=-19/5+(12/5)i}}} which is in standard form {{{a+bi}}} where {{{a=-19/5}}} and {{{b=12/5}}}