Question 642162
{{{3y^2=20}}}


{{{3y^2-20=0}}}


{{{3y^2+0y-20=0}}}


Now use the quadratic formula to solve for y


In this case, a = 3, b = 0, c = -20, 


Plug all this into the quadratic formula to get


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


{{{y = (-(0)+-sqrt((0)^2-4(3)(-20)))/(2(3))}}}


{{{y = (0+-sqrt(0-(-240)))/(6)}}}


{{{y = (""+-sqrt(0+240))/(6)}}}


{{{y = (""+-sqrt(240))/6}}}


{{{y = (sqrt(240))/6}}} or {{{y = (-sqrt(240))/6}}}


{{{y = (4*sqrt(15))/6}}} or {{{y = (-4*sqrt(15))/6}}}


{{{y = (2*sqrt(15))/3}}} or {{{y = (-2*sqrt(15))/3}}}


{{{y = 2.58198889747161}}} or {{{y = -2.58198889747161}}}


Note: The last solutions above are approximate