(2х^2+3х)^2-7(2х^2+3х)=-10

$(2x^2+3x)^2-7(2x^2+3x)=-10 \\ t = 2x^2+3x \\ t^2-7t+10=0 \\ D = 49-4*10 = 9 \\ t_{1,2} = \frac{7 \pm 3}{2} \\ t_1 = 5, t_2 = 2 \\ 2x^2+3x-5=0 \\ D = 3^2+5*4*2 = 49 \\ x_{1,2} = \frac{-3 \pm 7}{4} \\ x_1 = 1; x_2 = -\frac{5}{2} \\ 2x^2+3x-2=0 \\ D = 9+4*2*2 = 25 \\ x_{3,4} = \frac{-3 \pm 5}{4} \\ x_3 = -2; x_4 = \frac{1}{2} \\ x \in \{-\frac{5}{2}; -2; \frac{1}{2}; 1 \} \\ (x-1)(x-\frac{1}{2})(x+2)(x+\frac{5}{2}) = 0 \\$