Question 627162
{{{3u^2+10u=-3}}}


{{{3u^2+10u+3=0}}}


Now use the quadratic formula to solve for 'u':


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


{{{u = (-(10)+-sqrt((10)^2-4(3)(3)))/(2(3))}}}


{{{u = (-10+-sqrt(100-(36)))/(6)}}}


{{{u = (-10+-sqrt(64))/6}}}


{{{u = (-10+sqrt(64))/6}}} or {{{u = (-10-sqrt(64))/6}}}


{{{u = (-10+8)/6}}} or {{{u = (-10-8)/6}}}


{{{u = -2/6}}} or {{{u = -18/6}}}


{{{u = -1/3}}} or {{{u = -3}}}


So the solutions are {{{u = -1/3}}} or {{{u = -3}}}