Answer
No solution.
Work Step by Step
Step 1. Start from the original equation and form squares with the variables: $x^2+20x+4y^2-40y=-300$ or $(x^2+20x+100)+4(y^2-10y+25)=100+100-300$ which gives $(x-10)^2+4(y-5)^2=-100$
Step 2. Clearly the above equation has no solutions as both squares are larger or equal to zero.
Step 3. We can not graph the equation.
