Verified answer

[tex]\displaystyle 1)x(x+9)=0\\x=0,x+9=0\\x_1=-9,x_2=0\\\\2)(x-8)(x+3,5)=0\\\\(x-8)(x+\frac{7}{2})=0\\ \\x-8=0,x+\frac{7}{2}=0\\ \\x_1=-\frac{7}{2}=-3,5; x_2=8\\ \\3)(3x+2,7)(4x-13)=0\\\\(3x+\frac{27}{10})(4x-13)=0\\ \\3x+\frac{27}{10}=0,4x-13=0\\ \\x_1=-\frac{9}{10}=-0,9;x_2=\frac{13}{4}=3,25[/tex]