УРАВНЕНИЕ! (2x^2 + 7x - 8) (2x^2 + 7x - 3) -6 =0

Решите задачу:

$(2x^2 + 7x - 8) (2x^2 + 7x - 3) -6 =0 \\ 2x^2+7x-3=t \\ t(t-5)-6=0 \\ t^2-5t-6=0 \\ D=25-4*2*(-6)=49 \\ t_{1,2} = \frac{5б7}{2} \\ t_1 = -1 ; t_2 = 6 \\ 2x^2+7x-3=-1 \\ 2x^2+7x-2=0 \\ D=49-2*4*(-2)=65 \\ x_{1,2} = \frac{-7б \sqrt{65} }{4} \\ 2x^2+7x-3=6 \\ 2x^2+7x-9=0 \\ D=49-4*2*(-9)=121 \\ x_{3,4} = \frac{-7б11}{4} \\ x_{3} = 1 ; x_4 = -4,5 \\$