求几道一元二次方程解法3x^2-2x-4=0x^2-7/2x-2=0(x^2-1)/2-(x-2)/3=x^2(x-1)^2-7(x-1)-8=0

Question

求几道一元二次方程解法
3x^2-2x-4=0
x^2-7/2x-2=0
(x^2-1)/2-(x-2)/3=x^2
(x-1)^2-7(x-1)-8=0

Answer 1

3x^2-2x-4=0
x1=(2+(4+48)^0.5)/6=(1+13^0.5)/3
x2=(2-(4+48)^0.5)/6=(1-13^0.5)/3
x^2-7/2x-2=0
2x^2-7x-4=0
(2x+1)(x-4)=0
x1=-1/2
x2=4
(x^2-1)/2-(x-2)/3=x^2
3(x^2-1)-2(x-2)=6x^2
3x^2-3-2x+4=6x^2
3x^2+2x-1=0
(3x-1)(x+1)=0
x1=1/3
x2=-1
(x-1)^2-7(x-1)-8=0
(x-1+1)(x-1-8)=0
x*(x-9)=0
x1=0
x2=9