Solving quadratic when c = 0

A quadratic of the form $ax^2 + bx = 0$ — with the constant term $c = 0$ — is the easiest kind to solve.
Both terms share a factor of $x$, so you can factor it out instead of reaching for the quadratic formula.

Factoring out $x$ turns $x^2 - 5x = 0$ into $x(x - 5) = 0$.
A product equals zero only when one of its factors is zero, so either $x = 0$ or $x - 5 = 0$.
This gives the two solutions $x = 0$ and $x = 5$.

Notice that one solution is always $x = 0$ when $c = 0$.
That is because $x$ is a factor of the whole expression, so setting that factor to zero is always allowed.
The other solution comes from setting the bracket equal to zero.

This also works when the $x$-term has a coefficient.
For $0.2x(40 - x) = 0$ the expression is already factored, so $x = 0$ or $40 - x = 0$, giving $x = 0$ or $x = 40$.
These are often the points where a curve crosses the $x$-axis.

Solve $x^2 - 7x = 0$.
Factor out $x$: $x(x - 7) = 0$.
So $x = 0$ or $x - 7 = 0$, giving $x = 0$ or $x = 7$.

At which points does the line $y = x$ meet the parabola $y = x^2$?
Setting them equal, $x^2 = x$, so $x^2 - x = 0$ and $x(x - 1) = 0$.
Then $x = 0$ or $x = 1$, giving the points $(0, 0)$ and $(1, 1)$.

Solve $3x^2 + 12x = 0$.
Factor out $3x$: $3x(x + 4) = 0$.
So $3x = 0$ or $x + 4 = 0$, giving $x = 0$ or $x = -4$.