#### Answer

$1-2i$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
\dfrac{-12-i}{-2-5i}
,$ multiply the numerator and the denominator by the conjugate of the denomniator. Use special products to multiply the resulting expression. Then use the equivalence $i^2=-1.$ Finally, combine the real parts and the imaginary parts.
$\bf{\text{Solution Details:}}$
Reversing the operator between the terms of the expression, $
-2-5i
,$ then its conjugate is $
-2+5i
.$
Multiplying the numerator and the denominator of the given expression by the conjugate of the denominator, the given expression is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{-12-i}{-2-5i}\cdot\dfrac{-2+5i}{-2+5i}
\\\\=
\dfrac{(-12-i)(-2+5i)}{(-2-5i)(-2+5i)}
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel}
\dfrac{-12(-2)-12(5i)-i(-2)-i(5i)}{(-2-5i)(-2+5i)}
\\\\=
\dfrac{24-60i+2i-5i^2}{(-2-5i)(-2+5i)}
.\end{array}
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\dfrac{24-60i+2i-5i^2}{(-2)^2-(5i)^2}
\\\\=
\dfrac{24-60i+2i-5i^2}{4-25i^2}
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{24-60i+2i-5(-1)}{4-25(-1)}
\\\\=
\dfrac{24-60i+2i+5}{4+25}
.\end{array}
Combining the real parts and the imaginary parts results to
\begin{array}{l}\require{cancel}
\dfrac{(24+5)+(-60i+2i)}{4+25}
\\\\=
\dfrac{29-58i}{29}
\\\\=
\dfrac{\cancel{29}-\cancel{29}(2)i}{\cancel{29}}
\\\\=
1-2i
.\end{array}