Given f(x)=x2+3x-5 on [-1,3]
show f(x)= 10 without solving for x
f(x) is not 10
Not sure what you're after here. Do you mean that there exists a c such that -1 <= c <= 3 where f(c) = 10?
f(-1) = 3
f(3) = 13
Since f(x) is continuous, there is such a c.