Solve x^2 - 4y + 1 = 0 for y.

Step-by-step solution

1. 1

   Note first that $x^{2}-4y+1$ on its own is an expression, not an equation. With no equals sign there is nothing to solve and nothing to simplify — the three terms $x^{2}$, $-4y$ and $1$ are unlike, so they cannot be combined. Setting it equal to $0$ is what turns it into a solvable relation, and the standard request is to make $y$ the subject.

2. 2

   Isolate the $y$ term by moving everything else across. Subtract $x^{2}$ and $1$ from both sides:

   $-4y=-x^{2}-1$

   Keeping $-4y$ intact for now (rather than splitting off the minus) avoids a sign slip in the next step.

3. 3

   Divide both sides by $-4$. Dividing by a negative number changes the sign of every term on the right:

   $y=\frac{-x^{2}-1}{-4}=\frac{x^{2}+1}{4}$

   Equivalently $y=\tfrac14x^{2}+\tfrac14$. If only one of the two signs is flipped the result is wrong — a good reason to divide the whole fraction at once.

4. 4

   Check by substituting back. With $y=\frac{x^{2}+1}{4}$, the left side becomes

   $x^{2}-4\cdot\frac{x^{2}+1}{4}+1=x^{2}-(x^{2}+1)+1=0\ \checkmark$

   which holds for every $x$, as it must. Numerically at $x=-3$, $0$ and $2.5$ the residual is $0$ to machine precision ✓.

5. 5

   Read the geometry. $y=\tfrac14x^{2}+\tfrac14$ is a parabola opening upward, with vertex at $\left(0,\tfrac14\right)$ and a horizontal stretch by a factor of $2$ relative to $y=x^{2}$. Because $y$ appears only to the first power, the original equation defines $y$ as a genuine function of $x$ — solving for $x$ instead would give $x=\pm\sqrt{4y-1}$, which is not a function.