The quadratic equation has roots x=2/3 and x=-4. Find one set of possible values for a,b,c?

We'll put the quadratic as ax^2 + bx + c =
0

Now, we'll write the quadratic as a product of linear
factors:

ax^2 + bx + c = (x -
2/3)(x+4)

We'll remove the brackets from the right
side:

ax^2 + bx + c = x^2 + 4x - 2x/3 -
8/3

We'll compare both sides and we'll
get:

a =
1

b = 4 -
2/3

b =
10/3

c =
-8/3

Wealso could use Viete's relations to
determine the coefficients a,b,c:

x1 + x2 =
-b/a

But, from enunciation, x1 = 2/3 and x2 =
-4:

2/3 - 4 = -b/a

We'll
multiply by 3:

-10/3 =
-b/a

b/a =
10/3

10a =
3b

x1*x2 =
c/a

-8/3 =
c/a

-8a =
3c

The quadratic equation
is:

ax^2 + bx + c = x^2 +
10x/3 - 8/3