Question

I need help to solve it, be serious if you don't know don't answer please.

Answer 1

`k^2-8k+20=8`
`k^2-8k+12=0`
`k^2-8k=-12`
`k^2-8k+16=-12+16`
`(k-4)^2=4`

then k-4=2 or k-4=-2

and we have that k=6 and k=2, are the two answers for the equation.

if we replace k=2 in the first equation

`(2)^2-8(2)+20=4-16+20=-12+20=8`

which makes the equation true for k=2, then is we replace k=6 in the first equation:

`(6)^2-8(6)+20=36-48+20=-12+20=8`

then the equation holds for k=6 too.

the second equation is:

`2=\sqrt[]{-4-x}`

then

`4=-4-x`
`x=-4-4`

finally

`x=-8`

if we replace x=-8 in the second equation

`\sqrt[]{-4-(-8)}=\sqrt[]{-4+8}=\sqrt[]{4}=2`

then the equation holds for x=-8.