Verified answer

[tex]\displaystyle 3(x+4)^2=10x+32\\3(x^2+8x+16)=10x+32\\3x^2+24x+48=10x+32\\3x^2+24x+48-10x-32=0\\3x^2+14x+16=0\\3x^2+8x+6x+16=0\\x(3x+8)=2(3x+8)=0\\(3x+8)(x+2)=0\\3x+8=0,x+2=0\\\\x_1=-\frac{8}{3},x_2=-2[/tex]