Consider this system of equations: y = x2 y = x + k for which value of k does the system have no real number solutions? -2 for which value of k does the system have one real number solution? -0.25 for which value of k does the system have two real number solutions?   2

We know that

In a quadratic equation of the form
ax²+bx+c=0
the discriminant is equal to
(b²-4ac)
if (b²-4ac) > 0 then the system has two real number solutions
if (b²-4ac) =0 then the system has one real number solution
if (b²-4ac) < 0 then the system has no real number solutions

we have
y = x²  and  y = x + k

to resolve the system equate both equations
x²=x+k-----------> x²-x-k=0
a=1
b=-1
c=-k
find the disciminant
(b²-4ac)=(-1)²-4*(1)*(-k)-----> 1+4k

Part 1)For which value of k does the system have no real number solutions?
we know that
if (b²-4ac) < 0 then the system has no real number solutions
so
(1+4k) < 0-------> 4k < -1
k< -0.25

the answer part 1) is 
for k < -0.25 the system has no real number solutions
example
 if k=-2
-2 < -0.25-----> is ok
the system has no real number solutions
see the attached figure N 1

Part 2) For which value of k does the system have one real number solution?
we know that
if (b²-4ac) =0 then the system has one real number solution
so
(1+4k) = 0-------> k=-0.25

the answer part 2) is
for k=-0.25 the system has one real number solution
see the attached figure N 2

Part 3) For which value of k does the system have two real number solutions?
we know that
if (b²-4ac) > 0 then the system has two real number solutions
so
(1+4k) > 0------> 4k > -1
k > -0.25

the answer Part 3) is 
for k > -0.25 the system has two real number solutions
example
if k=2
2 > -0.25-----> is ok
the system has two real number solutions
see the attached figure N 3

Is there more to the question?