I have no idea how to do this!!!

Let ΔABC be a triangle such that b=sqrt(3), c=sqrt(10), and cos(B)=3/sqrt(10). Find all possible values for side length a.

Please show work

Law of cosines:

a^2=b^2+c^2 -2acCosB
=3+10-2sqrt(3*10)*3/sqrt10
=13-6sqrt3
finish it for solving a.
check my math.

sinB = 1/√10

b/sinB = c/sinC, so
√3/(1/√10) = √10/sinC
sinC = 1/√3

A = 180 - (B+C)
sinA = (180 - (B+C)) = sin(B+C)
= sinBcosC+cosBsinC
= (1/√10)(√2/√3)+(3/√10)(1/√3)
= (2+3√2)/(2√15)

Now get
a/sinA = b/sinB
or
a^2 = b^2+c^2 - 2bc*cosA

a^2=b^2+c^2-2bc*cosA

and we don't know A yet

To find all possible values for side length a, we can use the Law of Cosines, which states that in a triangle with sides a, b, and c, and opposite angles A, B, and C respectively, the following equation holds:

c^2 = a^2 + b^2 - 2ab * cos(C)

In our given triangle ΔABC, we are given values for b, c, and cos(B). Substituting these values into the Law of Cosines, we get:

10 = a^2 + 3 - 2a * (3 / sqrt(10))

Now, let's simplify the equation:

10 = a^2 + 3 - (6 / sqrt(10)) * a

Rearrange and group the terms:

a^2 - (6 / sqrt(10)) * a + 3 - 10 = 0

a^2 - (6 / sqrt(10)) * a - 7 = 0

Now, we have a quadratic equation in terms of a. To solve this equation, we can either factorize it or use the quadratic formula.

Let's use the quadratic formula:

a = (-b ± sqrt(b^2 - 4ac)) / (2a)

Here, a = 1, b = -(6 / sqrt(10)), and c = -7.

Substituting these values into the quadratic formula, we get:

a = (-(-(6 / sqrt(10))) ± sqrt((-(6 / sqrt(10)))^2 - 4 * 1 * -7)) / (2 * 1)

This simplifies to:

a = (6 / sqrt(10) ± sqrt((36 / 10) + 28)) / 2

a = (6 / sqrt(10) ± sqrt(64 / 10)) / 2

a = (6 / sqrt(10) ± 8 / sqrt(10)) / 2

Now, let's simplify further:

a = (6 + 8) / (2 * sqrt(10)) or a = (6 - 8) / (2 * sqrt(10))

a = 14 / (2 * sqrt(10)) or a = -2 / (2 * sqrt(10))

Simplifying gives:

a = 7 / sqrt(10) or a = -1 / sqrt(10)

Therefore, the possible values for side length a are a = 7 / sqrt(10) or a = -1 / sqrt(10).