Question 7622
{{{y=sqrt(3y+10)}}}
Squaring both sides we get
{{{y^2 = (sqrt(3y+10))^2}}}
{{{Y^2= 3y+10}}}
{{{y^2 - 3y -10=0}}}
This is a quadratic equation and we can solve this by usingthe formulae
{{{-b+-sqrt(b^2-4ac)/2a}}}
where
a=1,b=-3 and c=-10
Substituting values in the formulae
{{{3+-sqrt((-3)^2-4*1*-10)/2}}}
{{{(3+- 7)/2}}}
y=5,{{{-1/2}}}