Find a,b,c x^2+4y^2+16x-80y+320=0

$x^{2} + 4 y^{2} + 16 x - 80 y + 320 = 0$

Once the quadratic is in standard form, the values of $a$ , $b$ , and $c$ can be found.

$a x^{2} + b x + c$

Use the standard form of the equation to find $a$ , $b$ , and $c$ for this quadratic.

$a = 1$ , $b = 16$ , $c = 4 y^{2} - 80 y + 320$