Question 74406
let {{{y=sqrt(x)}}}, then substitute

{{{2y^2-7y-30=0}}}

using quadratic formula:  {{{y=(7+-sqrt(289))/4}}}  or  {{{y=(7+-17)/4}}}

so  {{{y=6}}}  and  {{{y=-(5/2)}}}

since {{{x=y^2}}};

then  {{{x=36}}}  and  {{{x=25/4}}}


you should note that for the {{{x=25/4}}} value, it is the negative square root that satisfies the equation