Question 617512


{{{(5y+3)^2+144=0}}} Start with the given equation.



{{{(5y+3)^2=0-144}}}Subtract {{{144}}} from both sides.



{{{(5y+3)^2=-144}}} Combine like terms.



{{{5y+3=""+-sqrt(-144)}}} Take the square root of both sides.



{{{5y+3=sqrt(-144)}}} or {{{5y+3=-sqrt(-144)}}} Break up the "plus/minus" to form two equations.



{{{5y+3=12i}}} or {{{5y+3=-12*i}}}  Simplify {{{sqrt(-144)}}} to get {{{12i}}}.



{{{5y=-3+12i}}} or {{{5y=-3-12*i}}} Subtract {{{3}}} from both sides.



{{{y=(-3+12i)/5}}} or {{{y=(-3-12i)/5}}} Divide both sides by 5