Question 824355
3(y^2+3)= 28y


3y^2+9 = 28y


3y^2+9-28y = 0


3y^2-28y+9 = 0


Use the quadratic formula to solve for y


{{{y = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{y = (-(-28)+-sqrt((-28)^2-4(3)(9)))/(2(3))}}} Plug in {{{a = 3}}}, {{{b = -28}}}, {{{c = 9}}}


{{{y = (28+-sqrt(784-(108)))/(6)}}}


{{{y = (28+-sqrt(676))/6}}}


{{{y = (28+sqrt(676))/6}}} or {{{y = (28-sqrt(676))/6}}}


{{{y = (28+26)/6}}} or {{{y = (28-26)/6}}}


{{{y = 54/6}}} or {{{y = 2/6}}}


{{{y = 9}}} or {{{y = 1/3}}}



The two solutions are


{{{y = 9}}} or {{{y = 1/3}}}