Divide the quintic by 4: x^5+7x^4-21x^3/4-147x^2/4+27x/4+189/4. The factors of 189 are 3*3*3*7. So we could have as factors 3, 3/2, 3/4, 7, 7/2, 7/4. These are tentative solutions for x without taking account of the sign. To get + from 5 factors we have five pluses, 3 pluses and 2 minuses, 1 plus and 4 minuses. But we have negative coefficients, so we can rule out 5 pluses. We need to trial positive and negative factors.
It seems unlikely that +7 is a factor because we have 4*7^5 and 28*7^4 governing the function. 28*7^4/147*7^2=4*7^2/21 which is a factor of about 10, showing that even 147x^2 is too small to have any significant effect on 28x^4. It seems more probable that -7 will be more effective as the odd powers of -7 will counteract the even powers to some extent, so -7 is a possible solution.
We also have 7/2 and 7/4 to consider, both positive and negative, as well as 3, 3/2, 3/4.
Synthetic division is probably the easiest way to check factors. Let's use it to check for -7:
-7 | 4..28 -21 -147 27 189
......4 -28....0..147...0 -189
......4....0 -21......0 27 |....0
So x+7 is a factor and we have a quartic: 4x^4-21x^2+27. However, this is effectively a quadratic, because there are no odd powers. We quickly realise that the factors of 27 we need are 3 and 9, and the factors of 4 are 4 and 1 because 4*3+1*9=21: (4x^2-9)(x^2-3)=(2x-3)(2x+3)(x-sqrt(3))(x+sqrt(3)).
Therefore the solution is: x=-7, 3/2, -3/2, sqrt(3), -sqrt3).