Can you help on a math problem

(x+3)(x+1) = -6(x-1) + 8

explain?

Expand it out to get:
(x+3)(x+1) = x^2 + 3x + x + 3
6(x-1) = 6x - 6

Bring them all to the same side to get:
(x+3)(x+1) + 6(x-1) - 8 = 0

Use the expansion that you have worked out to get:
x^2 + 4x + 3 + 6x - 6 - 8 = 0
=> x^2 + 10x - 11 = 0

Now you can solve for it using the usual method for obtaining the roots of a quadratic equation. Alternatively, you can use the following concept:
Quadratic equation : x^2 - (sum of roots)*x + (product of the roots) = 0

So in the current case, we have:
sum of roots = -10
product of roots = -11

The two numbers that satisfy this are -11 and 1. So use this to simplify:

x^2 - 11x + x - 11 = 0
=> x(x-11) + 1(x-11) = 0
=> (x-11)*(x+1) = 0
=> x = 11 or x = -1

Simplify first: x^2+4x+3=-6x+14

Now combine on one side: x^2+10x-11=0

Now factor: (x-1)(x+11)=0

Solve: x=1 or -11

Hoping this helps!