Complex roots or zeroes come in pairs when the coefficients of the polynomial are real. So three complex zeroes are not possible. I believe that this quintic has one pair of complex zeroes and three real zeroes. This can be proved be noting the change of sign of the polynomial when we substitute various values for x. The changes of sign can be seen:
y(-5)=-545, y(-4)=180; y(-1)=3, y(0)=-20; y(2)=-48, y(3)=103. The three changes of sign reveal 3 zeroes, (between x values: -5 and -4, -1 and zero, and 2 and 3) which leave two complex zeroes.
The two complex zeroes form a conjugate pair: a+ib and a-ib, which correspond to factors x-a-ib and x-a+ib. Multiply these factors and we get the real quadratic (x-a)2+b2=x2-2ax+a2+b2. Together with the three real factors we get the given polynomial.
We can use Newton's iterative method to find the three real zeroes, then we can derive the quadratic for the complex zeroes.
Newton's Method is xn+1=xn-y(xn)/y'(xn) where y' is the first derivative of y=5x4+8x3-30x2+4x-10.
x0 is chosen to start the process and x0 will take different values based on an approximation of the zeroes, which we can deduce from the y values calculated earlier. So I chose x0=-4, 0, and 2 as first approximations, and arrived at approximate zeroes:
-4.425175, -0.932973 and 2.588034.
We don't need high accuracy when we are certain that these are actual zeroes, because dividing by the corresponding factors has to be exact with no remainder. Progressive synthetic division can be used to get to the quadratic.
-4.425 | 1 2 -10 2 -10 -20
1 -4.425 10.7318 -3.2385 5.4804 | 20
-0.933 | 1 -2.425 0.7318 -1.2385 -4.5196 | 0
1 -0.933 3.1331 -3.6058 | 4.5196
2.588 | 1 -3.358 3.8649 -4.8443 | 0
1 2.588 -1.9927 |-4.8453
1 -0.770 1.8722 | 0 is the quadratic x2-0.77x+1.8722, from which x=(0.77±√-6.8959)/2.
x=½(0.77±2.626i)=0.385±1.313i approx.
To obtain a more accurate solution we can examine the derivation of the quadratic coefficients and replace them with more accurate figures.
