Solve by Factoring 1/4y^2+y=0

$\frac{1}{4} y^{2} + y = 0$

Factor $y$ out of $\frac{1}{4} y^{2} + y$ .

Factor $y$ out of $\frac{1}{4} y^{2}$ .

$y (\frac{1}{4} y) + y = 0$

Raise $y$ to the power of $1$ .

$y (\frac{1}{4} y) + y = 0$

Factor $y$ out of $y^{1}$ .

$y (\frac{1}{4} y) + y \cdot 1 = 0$

Factor $y$ out of $y (\frac{1}{4} y) + y \cdot 1$ .

$y (\frac{1}{4} y + 1) = 0$

Combine $\frac{1}{4}$ and $y$ .

$y (\frac{y}{4} + 1) = 0$

Write $1$ as a fraction with a common denominator.

$y (\frac{y}{4} + \frac{4}{4}) = 0$

Combine the numerators over the common denominator.

$y (\frac{y + 4}{4}) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y = 0$

$\frac{y + 4}{4} = 0$

Set $y$ equal to $0$ .

$y = 0$

Set $\frac{y + 4}{4}$ equal to $0$ and solve for $y$ .

Set $\frac{y + 4}{4}$ equal to $0$ .

$\frac{y + 4}{4} = 0$

Solve $\frac{y + 4}{4} = 0$ for $y$ .

Set the numerator equal to zero.

$y + 4 = 0$

Subtract $4$ from both sides of the equation.

$y = - 4$

The final solution is all the values that make $y (\frac{y + 4}{4}) = 0$ true.

$y = 0, - 4$