(Root 3+root 2)^x +(root 3-root 2)^x=10 solve for x please provide the total step

(√3 + √2)^x + (√3 - √2)^x = 10
(√3 + √2)^x*(√3 - √2)^x + ((√3 - √2)^x)^2 = 10(√3 - √2)^x
((√3 + √2)(√3 - √2))^x + (√3 - √2)^(2x) = 10(√3 - √2)^x
(3 - 2)^x + (√3 - √2)^(2x) = 10(√3 - √2)^x
1 + (√3 - √2)^(2x) = 10(√3 - √2)^x
(√3 - √2)^(2x) - 10(√3 - √2)^x + 1 = 0
((√3 - √2)^x)^2 - 10(√3 - √2)^x + 1 = 0
Set u = √3 - √2
u^2 - 10u + 1 = 0
u = 5 + 2√6
u = 5 - 2√6

(√3 - √2)^x = 5 + 2√6
x = log(5 + 2√6)/log(√3 - √2)
x = 2

(√3 - √2)^x = 5 - 2√6
x = log(5 - 2√6)/log(√3 - √2)
x = -2

The key to this one is to recognise that
(√3 + √2)(√3 - √2) =1
Therefore
(√3 - √2)^x = (√3 + √2)^(-x)

So if we write y = (√3 + √2)^x
the equation is

y + 1/y = 10
Multiply by y and then move all terms to the left side:

y² - 10y + 1 = 0

Either by completing the square, or by using the quadratic formula, we get
y = 5 ± 2√6
i.e. (√3 + √2)^x = 5 ± 2√6

Seeing √6 = √3 * √2 we immediately think of the fact that
(√3 + √2)² = 3 + 2√6 + 2 = 5 + 2√6 so x = 2 satisfies the equation.

Also since
(√3 - √2)^x = (√3 + √2)^(-x)
we know that
(√3 + √2)^(-2) = (√3 - √2)^2 = 5 - 2√6

Hnece the complete solution is
x = 2 or -2.