Find the value of K for which the following equations have 2 roots: x^2 -3x +k help please.?

The equation x² - 3x + k = 0 has two roots.

Then, discriminant b² - 4ac > 0
(-3)² - 4(1)(k) > 0
9 - 4k > 0
-4k > -9
4k < 9
k < 9/4

An equation should have = somewhere so I
assume you mean x^2-3x+k=0.
There is not just one value of k so I assume
you mean the set of values of k.
This is where b^2-4ac>0 for different real roots. a=1,b=-3, c=k so
9-4k>0 which gives k<9/4.

Could be several answers? Any more information?

a = 1
b = -3
c = k

b^2 - 4ac
If negative = imaginary
b^2 - 4ac becomes (-3)^2 - 4*1*k
= 9 - 4k
k = 9/4

If k greater than 9/4 then discriminant negative = imaginary

let k = 9/4

x^2 - 3x + 9/4
solve for roots
x = (-b +/- sqrt(b^2-4ac))/2a
x = -(9/4) +/- sqrt (9 - 4*1*(9/4)) / 2
x = (-9/4) +- sqrt(9 - 9) / 2 = real roots

Answer: k less than 9/4 = real roots

let c = 10/4
x = (-10/4) +/- sqrt(9 - 4*1*(10/4)) / 2
x = (-10/4) +/- sqrt(9 - 10) / 2
sqrt (9-10) is negative causing roots = imaginary.

Answer: k greater than 9/4 makes the root imaginary.