How would I put this parabola in standard form: x^2-6x+8y-31

Here how I started it:

x^2-6x+9=-8y+31

Im not sure what to do from here :(

If x^2-6x+8y-31 mean:

x^2-6x+8y-31=0 then

8y= -x^2+6x+31 Divide with 8

y= -x^2/8 + 6x/8 + 31/8

y= -x^2/8 + 3x/4 + 31/8

To put the given quadratic equation in standard form, which is in the form of "ax^2 + bx + c = 0," you need to rearrange the terms. Let's go step by step:

Start with the given equation: x^2 - 6x + 8y - 31 = 0

Step 1: Group the "x" terms together and the constant terms together.

(x^2 - 6x) + (8y - 31) = 0

Step 2: Complete the square for the "x" terms. To do this, take half of the coefficient of x (-6/2 = -3) squared, which is 9. Add and subtract 9 inside the first parentheses.

(x^2 - 6x + 9 - 9) + (8y - 31) = 0

Step 3: Rewrite the equation, taking the square from the first parentheses.

(x - 3)^2 - 9 + (8y - 31) = 0

Step 4: Combine like terms.

(x - 3)^2 +8y - 40 = 0

Step 5: Rearrange the equation to match the standard form.

(x - 3)^2 + 8y = 40

Therefore, the parabolic equation in standard form is (x - 3)^2 + 8y = 40.