According to ques,

\[~{{x}^{2}}~\text{ }11/4\text{ }x\text{ }+\text{ }15/8\text{ }=\text{ }0\]

Taking L.C.M ,

\[\left( 8{{x}^{2}}~\text{ }22x\text{ }+\text{ }15 \right)/\text{ }8\text{ }=\text{ }0\]

Or,

\[8{{x}^{2}}~\text{ }22x\text{ }+\text{ }15\text{ }=\text{ }0\]

Or,

\[8{{x}^{2}}~\text{ }12x\text{ }\text{ }10x\text{ }+\text{ }15\text{ }=\text{ }0\]

Or,

\[4x\left( 2x\text{ }\text{ }3 \right)\text{ }\text{ }5\left( 2x\text{ }\text{ }3 \right)\text{ }=\text{ }0\]

Or,

\[\left( 4x\text{ }\text{ }5 \right)\left( 2x\text{ }\text{ }3 \right)\text{ }=\text{ }0\]

Or,

So, \[4x\text{ }\text{ }5\text{ }=\text{ }0\text{ }or\text{ }2x\text{ }\text{ }3\text{ }=\text{ }0\]

Therefore,

\[x\text{ }=\text{ }5/4\text{ }or\text{ }x\text{ }=\text{ }3/2\]