Cardano’s method (simplified):

For x³ + ax² + bx + c

x = p/3u – u – a/3

where:

p = b – a²/3

u = ³√(q/2 ± √(q²/4 + p³/27))

and q = c + (2a³ – 9ab)/27

First, we need to determine p:

p = b – a²/3

p = 275 – (900/3) = 275 – 300 = -25

To determine u, we first need q:

q = c + (2a³ – 9ab)/27

q = -720 + (2(-27000) – (9(-30)(275))/27

q = -720 + (-54000 + 74250)/27

q = -720 + 750 = 30

Now, we can tackle u:

u = ³√(q/2 ± √(q²/4 + p³/27))

u = ³√(15 ± √(225 – 15625/27))

u = ³√(15 ± √((6075 – 15625)/27))

u = ³√(15 ± √(-9550)/27))

u = ³√(15 ± 18.807i)

Of course, now you’re stuck with taking the cubic root of an imaginary number, and that hurts my head.