Question

Please give an easy and fast way to solve this:

Answer 1

\[x ^{2}-xc-2c ^{2}=0\]

Answer 2

Use the quadratic formula.

Answer 3

\[(a+b^{2}=a ^{2} + 2ab + b ^{2}\]

Answer 4

solve what?

Answer 5

but isn't' a quadratic equation of the type:
\[a ^{2}+2ab+b ^{2}=0\]
OR
\[a ^{2}-2ab+b ^{2}=0\]
My equation has two -ve signs.

Answer 6

oh

Answer 7

You have to get x = 2c

Answer 8

(x+a)*(x-b)

Answer 9

Please show the steps.

Answer 10

x^2−xc−2*c^2=0

(x+c)(x-2c)=0

Answer 11

\[x = {c \pm \sqrt{c^2 - 4*z*(-2c^2)} \over 2}\]

Answer 12

lol

Answer 13

x^2-x c-2c^2=0

x=-c+-sqrt(c^2-4*1*2c^2)/2*1

Answer 14

OK. Thank You.

Answer 15

c can be treated as a constant

Answer 16

\[x^{2}-xc-c ^{2}=x^{2}-2xc+xc-c ^{2}=(x-2c)(x+c)=0 so we have x=2c or x=-c\]