According to the given question,

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

One of the roots is\[~{\scriptscriptstyle 3\!/\!{ }_4},\]and it satisfies the given equation

So,

\[8{{\left( 3/4 \right)}^{2~}}+\text{ }m\left( 3/4 \right)\text{ }+\text{ }15\text{ }=\text{ }0\]

Or,

\[8\left( 9/16 \right)\text{ }+\text{ }m\left( 3/4 \right)\text{ }+\text{ }15\text{ }=\text{ }0\]

Or,

\[18/4\text{ }+\text{ }3m/4\text{ }+\text{ }15\text{ }=\text{ }0\]

Taking L.C.M, we have

\[\left( 18\text{ }+\text{ }3m\text{ }+\text{ }60 \right)/4\text{ }=\text{ }0\]

Or,

\[18\text{ }+\text{ }3m\text{ }+\text{ }60\text{ }=\text{ }0\]

Or,

\[3m\text{ }=\text{ }\text{ }78\]

Or,

\[m\text{ }=\text{ }-26\]

By putting the value of \[m\], we get

\[8{{x}^{2~}}+\text{ }\left( -26 \right)x\text{ }+\text{ }15\text{ }=\text{ }0\]

Or,

\[8{{x}^{2~}}\text{ }26x\text{ }+\text{ }15\text{ }=\text{ }0\]

Or,

\[8{{x}^{2}}~\text{ }20x\text{ }\text{ }6x\text{ }+\text{ }15\text{ }=\text{ }0\]

Or,

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

Or,

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

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

Hence,

\[x\text{ }=\text{ }{\scriptscriptstyle 3\!/\!{ }_4}\text{ }or\text{ }x\text{ }=\text{ }5/2\]