How to find the zeros using vertex form?

-x²+3x+10 = 0
x²-3x-10 = 0
x²-3x-10 = 10
x²-3x+(3/2)²-(3/2)²-10 = 0
(x-3/2)²-12.25 = 0
(x-3/2)² = 12.25
|x-3/2| = 3.5
x-1.5 = 3.5 or -x+1.5 = 3.5

x = 5 or -2

-(x^2 - 3x + 10) = -(x-5)*(x+2) --, so zeroes occur at x = 5 and x = -2

y =  -x^2 + 3x + 10
y = -(x - 3/2)^2 + 49/4 = 0
(x-3/2)^2 = 49/4
x - 3/2 = ± 7/2 => can you take it from here?

Since f(x) is a polynomial, its domain is {real numbers}.

Since the parabola opens downward, it has a maximum value.  The maximum value is 12.25 or 49/4.  The range is {y: y ? 49/4}.

to find zeros means to find values of x that will make the equation zero.so you need to factor the equation to find the values of x. that would be the easiest way
-x^2+3x+10=0
x^2-3x-10=0
(x+5)(x-2)=0
therefor x= -5, 2
those values would make your equation zero and they are the zeros.

You have the answers for the zeros.

The domain is the set of real numbers except for any that are not allowed by any of the operations.

There's no restriction on x^2 (you can square any real number), or on 3x, or on adding -x^2 + 3x + 10. Those are all legitimate for all real numbers. So there is no restriction on the domain.

As for the range, a parabola either opens upward, in which case it has a minimum but goes up to +infinity, or it opens downward, in which case it has a maximum and goes down to -infinity.

This one opens downward since the coefficient on x^2 is negative. You have the vertex form so that tells you where the maximum is, which is at the vertex.