How do you find the standard form of x^2+y^2+8y+4x-5=0?

This is a circle ${\left(x - \left(- 2\right)\right)}^{2} + {\left(y - \left(- 4\right)\right)}^{2} = {5}^{2}$

Explanation:

From the given ${x}^{2} + {y}^{2} + 8 y + 4 x - 5 = 0$

Perform completing the square method

${x}^{2} + {y}^{2} + 8 y + 4 x - 5 = 0$

by rearranging the terms:

${x}^{2} + 4 x + {y}^{2} + 8 y - 5 = 0$

Calculate the numbers to be added on both sides of the equation

from $4 x$, take the 4, divide this number by 2 then square the result.
result $= 4$

from $8 y$, take the 8, divide this number by 2 then square the result.
result $= 16$

Therefore Add $4$ and $16$ to both sides of the equation and also transpose $- 5$ to the right side.

${x}^{2} + 4 x + 4 + {y}^{2} + 8 y + 16 - 5 = 0 + 4 + 16$

$\left({x}^{2} + 4 x + 4\right) + \left({y}^{2} + 8 y + 16\right) = 5 + 4 + 16$

${\left(x + 2\right)}^{2} + {\left(y + 4\right)}^{2} = 25$

From the standard form:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

so that

${\left(x - - 2\right)}^{2} + {\left(y - - 4\right)}^{2} = {5}^{2}$ the required standard form.

Have a nice day !!! from the Philippines..