how Do I find the X and Y intercepts of this equation? X(t)= -5x^2+105x+100?

y = -5x² + 105x + 100

When x = 0:
y = -5(100)² + 105(0) + 100
y = 0
Hence, the y-intercept = 100

When y = 0:
0 = -5x² + 105x + 100
x² - 21x - 20 = 0
x = [21 ± √(21² + 4*20)]/2
x = (21 + √521)/2 or x = (21 - √521)/2
The x-intercepts = (21 + √521)/2 and x = (21 - √521)/2

The way you wrote it, X(t) looks wrong. Do you mean y?
The y intercept is when x = 0. So just put in 0 instead of 'x', and figure out the value of y.
The x intercept is when y = 0. So solve 0 = -5x^2 + 105x + 100. You do remember the quadratic equation solution, don't you?

f(x) = -5x^2 + 105x + 100, x and y intercepts are ?

x = 0, f(x) = ?
f(0) = 0 + 0 + 100
f(0) = 100
y-intercept is 100.

f(x) = 0, x = ?
-5x^2 + 105x + 100 = 0
x^2 - 21x - 20 = 0
x = (21 ± √(521))/2
x-intercept are (21 - √(521))/2, (21 +√(521))/2.

y=-5x^2+105x+100
y=0=>
5x^2-105x-100=0=>
x=[105+/-sqr(-105)^2-4(5)(-100)]/[2(5)]=>
the x-intercept=21.9 or -.9 approximately.

x=0=>
the y-intercept=100