Answer:
1.
It has two solution because it is in quadratic form.
2.
(x-4)²-28=8
x²-8x+16-28-8=0
x²-8x-20=0
Doing middle term factorization
x²-(10-2)x-20=0
x²-10x+2x-20=0
x(x-10)+2(x-10)
(x-10)(x+2)=0
either
x=10
or
x=-2
:. x= 10 or -2
3.
2x²-5x-5=0
Comparing above equation with ax²+bx +c
we get
a=2
b=-5
c=-5
quadratic equation is
[tex]x = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a} [/tex]
Substituting value
[tex]x = \frac{ 5 ± \sqrt{ {-5}^{2} - 4×2×-5} }{2×-5} [/tex]
-10x=5±[tex] \sqrt{ {25+ 40}}[/tex]
taking + ve
-10x=5+[tex] \sqrt{ {65}}[/tex]
x=-½-[tex] \frac {\sqrt{ {65}}}{10}[/tex]
taking -ve
x=-½+[tex] \frac {\sqrt{ {65}}}{10}[/tex]
