To find equation solutions, solve $x=0$ and $x-5=0$.
$$x=0$$ $$x=5$$
Steps Using the Quadratic Formula
Subtract $5x$ from both sides.
$$x^{2}-5x=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-5$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{5±5}{2}$ when $±$ is plus. Add $5$ to $5$.
$$x=\frac{10}{2}$$
Divide $10$ by $2$.
$$x=5$$
Now solve the equation $x=\frac{5±5}{2}$ when $±$ is minus. Subtract $5$ from $5$.
$$x=\frac{0}{2}$$
Divide $0$ by $2$.
$$x=0$$
The equation is now solved.
$$x=5$$ $$x=0$$
Steps for Completing the Square
Subtract $5x$ from both sides.
$$x^{2}-5x=0$$
Divide $-5$, the coefficient of the $x$ term, by $2$ to get $-\frac{5}{2}$. Then add the square of $-\frac{5}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
