Ответ:

[tex]x_{1} = -2\frac{2}{3} \\x_{2} = 5[/tex]

Объяснение:

[tex]3x^2 - 7x - 40 = 0\\D = (-7)^2 - 4*3*(-40) = 49 + 480 = 529\\\sqrt{D} = \sqrt{529} = 23\\x_{1} = \frac{7 -23}{2*3} = -\frac{16}{6} = -2\frac{2}{3} \\x_{2} = \frac{7+23}{2*3} = \frac{30}{6} = 5[/tex]

[tex]\displaystyle 3x^2-7x-40=0\\3x^2+8x-15x-40=0\\x(3x+8)-5(3x+8)=0\\(3x+8)(x-5)=0\\3x+8=0,x-5=0\\\\x_1=-\frac{8}{3},x_2=5[/tex]