We should expect the two sections to be \[x\text{ }and\text{ }15\text{ }\text{ }x.\]
now,
\[150\text{ }=\text{ }45x\text{ }\text{ }3×2\]
\[3×2\text{ }\text{ }45x\text{ }+\text{ }150\text{ }=\text{ }0\]
Separating by 3, we get
\[x2\text{ }\text{ }15x\text{ }+\text{ }50\text{ }=\text{ }0\]
\[x2\text{ }\text{ }10x\text{ }\text{ }5x\text{ }+\text{ }50\text{ }=\text{ }0\]
\[x\left( x\text{ }\text{ }10 \right)\text{ }\text{ }5\left( x\text{ }\text{ }10 \right)\text{ }=\text{ }0\]
\[\left( x\text{ }\text{ }5 \right)\left( x\text{ }\text{ }10 \right)\text{ }=\text{ }0\]
Thus, \[x\text{ }\text{ }5\text{ }=\text{ }0\text{ }or\text{ }x\text{ }\text{ }10\text{ }=\text{ }0\]
\[x\text{ }=\text{ }5\text{ }or\text{ }10\]
In this way, in the event that one section is 5 the other part is 10 as well as the other way around.