It would be impossible to completely factor -12a2+20a-21 into the product of real numbers. The expression would be considered prime in that case. This is probably the answer you are looking for.
However, factoring would be possible if complex linear factors are allowed. It would require using the definition i = √(-1) and the quadratic formula:
a = (-B ± √(B2-4AC))/2A
My first step would be to factor out -1/3, turning A into a positive square number:
-12a2+20a-21 = -1/3(36a2-60a+63)
From there, I would use the quadratic formula with A=36, B=-60 and C=63:
a = (-(-60) ± √((-60)2-4(36)(63)))/(2(36))
= (60 ± √(3600-9072))/72 [Simplifying]
= (60 ± √(-5472))/72 [Subtracting]
= (60 ± 12i√(38))/72 [Simplifying the radical]
= 12(5 ± i√(38))/72 [Factoring the numerator]
= (5 ± i√(38))/6 [Reducing]
With a bit of rewriting this would give a factorization of:
-1/3(6a - 5 + i√(38))(6a - 5 - i√(38)) = -1/3(-6i a + √(38) + 5i)(6i a + √(38) - 5i)
Bonus:
It is also possible to complete the square using the expression -12a2+20a-21:
-12a2+20a-21 = -12(a2-5/3a)-21 [Rewriting]
= -12(a2-5/3a + 25/36)-21 + 25/3 [Completing the square]
= -12(a-5/6)2-63/3+25/3 [Rewriting]
= -12(a-5/6)2-38/3 [Simplifying]
This form is important, primarily because it shows that at a = 5/6 the expression reaches a maximum value of -38/3.