Question 187448
{{{5y^2=-40}}} Start with the given equation.



{{{5y^2+40=0}}} Add 40 to both sides.




Notice we have a quadratic equation in the form of {{{ay^2+by+c}}} where {{{a=5}}}, {{{b=0}}}, and {{{c=40}}}



Let's use the quadratic formula to solve for y



{{{y = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{y = (0 +- sqrt( (0)^2-4(5)(40) ))/(2(5))}}} Plug in  {{{a=5}}}, {{{b=0}}}, and {{{c=40}}}



{{{y = (0 +- sqrt( 0-4(5)(40) ))/(2(5))}}} Square {{{0}}} to get {{{0}}}. 



{{{y = (0 +- sqrt( 0-800 ))/(2(5))}}} Multiply {{{4(5)(40)}}} to get {{{800}}}



{{{y = (0 +- sqrt( -800 ))/(2(5))}}} Subtract {{{800}}} from {{{0}}} to get {{{-800}}}



{{{y = (0 +- sqrt( -800 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{y = (0 +- 20i*sqrt(2))/(10)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{y = (0)/(10) +- (20i*sqrt(2))/(10)}}} Break up the fraction.  



{{{y = 0 +- 2*sqrt(2)*i}}} Reduce.  



{{{y = 2*sqrt(2)*i}}} or {{{y = -2*sqrt(2)*i}}} Break up the expression.  



So the answers are {{{y = 2*sqrt(2)*i}}} or {{{y = -2*sqrt(2)*i}}}