solve for x (x+3)(3x+5)=7?

(x+3)(3x+5)=7 or (3x + 2)(x + 4) = 0 giving x = --4, -- 2/3

(x + 3)(3x + 5) = 7
3x^2 + 9x + 5x + 15 = 7
3x^2 + 14x + 15 - 7 = 0
3x^2 + 14x + 8 = 0
(3x + 2)(x + 4) = 0

3x + 2 = 0
3x = -2
x = -2/3

x + 4 = 0
x = -4

∴ x = -2/3 , -4

3x² + 14x + 8 = 0
(3x + 2)(x + 4) = 0
x = - 2/3 , x = - 4

expand the brackets,
leaving you wit 3x^2+11x+15=7,
take 7 to the other side of the equal sign by subtracting the LHS equation by 7,
leaving you wit, 3x^2+11x+8=0,
using the quadratic formula,
x=(-b) (+-) ((sqrt((b^2)-(4*a*c)))/2a)
(please if the quadratic equation is not clear to you pls do search for it)
then it will give you values for x.
(do corrct me if im wrong)

Algebraically only? Better use technology. Plot y = 3x^2 +14x +15 and y = 7/x and look for point of intersection, if it is x(x+3)(3x+5) =7.
If it is only (x+3)(3x+5) =7, use quadratic formula.

(x + 3) (3x + 5) = 7 ---> Simplify (x + 3) (3x + 5)
3x^2 + 5x + 9x + 15 = 7 ---> Simplified form
3x^2 + 14x + 15 = 7 ---> Add 5x and 9x together
3x^2 + 14x + 15 - 7 = 0 ---> - 7 to bring 7 to the left side
3x^2 + 14x + 8 = 0 ---> Subtract 7 from 15
(x + 4) (3x + 2) = 0 ---> Factorize 3x^2 + 14x + 8
(x + 4) = 0 OR (3x + 2) = 0
x = -4 OR x = -2/3 ---> Answer