Determine the quadratic ax^2+bx+c, if a,b,c are terms of arithmetic sequence. a=2t-3,b=5t+1,c=4t-7

If  2t-3, 5t+1, 4t-7, are the consecutive terms of an
arithmetical progression, then the middle term is the arithmetical mean of its neighbor
terms.

5t+1=(2t-3+4t-7)/2

We'll
combine like terms inside
brackets:

5t+1=(6t-10)/2

5t+1=2(3t-5)/2

5t+1=3t-5

We'll
move the terms in t to the left side and the numbers alone to the right
side:

5t-3t=-1-5

2t=-6

t=-6/2

t=-3

Since
the coefficients a,b,c are depending on t, we'll determine
them:

a = 2t - 3

a = -6 -
3

a = -9

b = 5t +
1

b = -15+1

b =
-14

c = 4t - 7

c = -12 -
7

c = -19

The quadratic
equation is: -9x^2 - 14x - 19 = 0

We'll multiply by -1 and
we'll get:

9x^2 + 14x + 19 =
0