help plz quick

Just to reformat your question for my aesthetic quick;

Simplify $(x+1)(2x+3)(3x+4) = (6x^{2}+5)(x-3)$ to the form $0 = Ax^{2}+Bx+C$ where A, B, and C are positive integers with a greatest common divisor of 1. What is $A+B+C$?

$(x+1)(2x+3)(3x+4) = (6x^{2}+5)(x-3)$

$(2x^2+3x+2x+3)(3x+4) = 6x^3-18x^2+5x-15$

$(2x^2+5x+3)(3x+4)=6x^3-18x^2+5x-15$

$6x^3+8x^2+15x^2+20x+9x+12=6x^3-18x^2+5x-15$

$6x^3+23x^2+29x+12=6x^3-18x^2+5x-15$

$41x^2+24x+27=0$

$Ax^2 + Bx + C = 0$

$A = 41, B = 24, C = 27$

$A + B + C = 41+24+27=92$