<P> In the case that b ≠ 0, there are two distinct roots, but if the polynomial is irreducible, they cannot be expressed in terms of square roots of numbers in the coefficient field . Instead, define the 2 - root R (c) of c to be a root of the polynomial x + x + c, an element of the splitting field of that polynomial . One verifies that R (c) + 1 is also a root . In terms of the 2 - root operation, the two roots of the (non-monic) quadratic ax + bx + c are </P> <Dl> <Dd> b a R (a c b 2) (\ displaystyle (\ frac (b) (a)) R \ left ((\ frac (ac) (b ^ (2))) \ right)) </Dd> </Dl> <Dd> b a R (a c b 2) (\ displaystyle (\ frac (b) (a)) R \ left ((\ frac (ac) (b ^ (2))) \ right)) </Dd> <Dl> <Dd> b a (R (a c b 2) + 1). (\ displaystyle (\ frac (b) (a)) \ left (R \ left ((\ frac (ac) (b ^ (2))) \ right) + 1 \ right).) </Dd> </Dl>

Given the formula d equals abc what is the formula for c