Discriminant = sq root of b^2 - 4*a*c =
sq root of 36 - 4*1*13 =
sq root 36 -52 =
sq root 36 -52 =
sq root -16
There are no real solutions for this equation. Both roots are complex numbers.
Answer:
Indefinite integration acts as a tool to solve many physical problems.
There are many type of problems that require an indefinite integral to solve.
Basically indefinite integration is required when we deal with quantities that vary spatially or temporally.
As an example consider the following example:
Suppose that we need to calculate the total force on a object placed in a non- uniform field.
As an example let us consider a rod of length L that posses an charge 'q' per meter length and suppose that we place it in a non uniform electric field which is given by

Now in order to find the total force on the rod we cannot use the similar procedure as we can see that the force on the rod varies with the position of the rod.
But if w consider an element 'dx' of the rod at a distance 'x' from the origin the force on this element will be given by

Now to find the whole force on the rod we need to sum this quantity over the whole length of the rod requiring integration, as shown

Similarly there are numerous problems considering motion of particles that require applications of indefinite integration.

solve for "k", to find k or the "constant of variation"
then plug k's value back to

now.... what is "p" when q = 5? well, just set "q" to 5 on the right-hand-side, and simplify, to see what "p" is
So there is an identity we'll need to use to solve this:
cos(x+y) = cosxcosy - sinxsiny
replace the numerator with the right hand side of that identity and we get:
(cosxcosy - sinxsiny)/cosxsiny
Separate the numerator into 2 fractions and we get:
cosxcosycosxsiny- sinxsiny/cosxsiny
the cosx's cancel on the left fraction, the siny's cancel on the right fraction and we're left with:
cosy/siny - sinx/cosx
which simplifies to:
coty - tanx