Find the Roots (Zeros) y=x^2+4x

$y = x^{2} + 4 x$

Set $x^{2} + 4 x$ equal to $0$ .

$x^{2} + 4 x = 0$

Solve for $x$ .

Factor $x$ out of $x^{2} + 4 x$ .

Factor $x$ out of $x^{2}$ .

$x \cdot x + 4 x = 0$

Factor $x$ out of $4 x$ .

$x \cdot x + x \cdot 4 = 0$

Factor $x$ out of $x \cdot x + x \cdot 4$ .

$x (x + 4) = 0$

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

$x = 0$

$x + 4 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $x (x + 4) = 0$ true.

$x = 0, - 4$