Answer
$y = \frac{2}{{625}}\sqrt {3 + 5t} \left( {25{t^2} - 20t + 24} \right) + C$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dt}} = \frac{{{t^2}}}{{\sqrt {3 + 5t} }} \cr
& {\text{Separate the variables}} \cr
& dy = \frac{{{t^2}}}{{\sqrt {3 + 5t} }}dt \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = \int {\frac{{{t^2}}}{{\sqrt {3 + 5t} }}} dt \cr
& y = \int {\frac{{{t^2}}}{{\sqrt {3 + 5t} }}} dt \cr
& {\text{Integrate by substitution}} \cr
& {\text{Let }}u = 3 + 5t,{\text{ }}t = \frac{{u - 3}}{5},{\text{ }}dt = \frac{1}{5}du \cr
& y = \int {{{\left( {\frac{{u - 3}}{5}} \right)}^2}\frac{1}{{\sqrt u }}} \left( {\frac{1}{5}} \right)du \cr
& y = \frac{1}{{125}}\int {\frac{{{u^2} - 6u + 9}}{{{u^{1/2}}}}} du \cr
& y = \frac{1}{{125}}\int {\left( {{u^{3/2}} - 6{u^{1/2}} + 9{u^{ - 1/2}}} \right)} du \cr
& y = \frac{2}{{625}}{u^{5/2}} - \frac{4}{{125}}{u^{3/2}} + \frac{{18}}{{125}}{u^{1/2}} + C \cr
& \cr
& {\text{Factoring}} \cr
& y = \frac{2}{{625}}{u^{1/2}}\left( {{u^2} - 10u + 45} \right) + C \cr
& {\text{Write in terms of }}x,{\text{ let }}u = 3 + 5t \cr
& y = \frac{2}{{625}}{\left( {3 + 5t} \right)^{1/2}}\left( {{{\left( {3 + 5t} \right)}^2} - 10\left( {3 + 5t} \right) + 45} \right) + C \cr
& y = \frac{2}{{625}}\sqrt {3 + 5t} \left( {9 + 30t + 25{t^2} - 30 - 50t + 45} \right) + C \cr
& y = \frac{2}{{625}}\sqrt {3 + 5t} \left( {25{t^2} - 20t + 24} \right) + C \cr} $$
