Transform 4y²+12y-3x+41=0 into standard form parabola?

4y^2+12y-3x+41 = 0

4(y^2+3y) = 3x-41
4(y^2+3y+(3/2)^2 = 3x-41 + 4(3/2)^2
4(y + 3/2)^2 = 3x-32
(y + 3/2)^2 = 3/4 (x - 32/3)

Did you know?

Did you know that a quadratic equation in the form of Ax² + Bx + C = 0 can be transformed into standard form to represent a parabola? To transform 4y² + 12y - 3x + 41 = 0 into standard form, we can rearrange the terms. By isolating the variable, the equation can be rewritten as 4y² + 12y = 3x - 41. The next step is completing the square on the quadratic terms. By adding (b/2)², where b is the coefficient of y, the equation becomes 4(y² + 3y + 9/4) = 3x - 41 + 4(9/4). Simplifying further, we have 4(y + 3/2)² = 3x - 41 + 9. Now, to achieve standard form, we divide both sides by 4, resulting in (y + 3/2)² = (3/4)x - 8.25. Therefore, through rearranging the equation and completing the square, we successfully transformed 4y² + 12y - 3x + 41 = 0 into standard form, representing a parabolic equation.