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The $\pm$ symbol indicates that the quadratic formula produces two distinct roots for $r$, denoted as $r_1$ and $r_2$. These two roots correspond to two independent exponential solutions, $e^{r_1 t}$ and $e^{r_2 t}$, which form the basis of the general solution.

Conditions: The characteristic equation is quadratic.; The discriminant $b^2 - 4ac$ is positive (yielding two distinct real roots in the context of this specific clip's introduction).

The $\pm$ symbol indicates that the quadratic formula produces two distinct roots for $r$, denoted as $r_1$ and $r_2$. These two roots correspond to two independent exponential solutions, $e^{r_1 t}$ and $e^{r_2 t}$, which form the basis of the general solution.

Conditions: The characteristic equation is quadratic.; The discriminant $b^2 - 4ac$ is positive (yielding two distinct real roots in the context of this specific clip's introduction).

The characteristic equation is the quadratic algebraic equation $0 = ar^2 + br + c$. It is obtained by substituting the trial solution $y = e^{rt}$ into the second-order linear homogeneous differential equation $ay'' + by' + cy = 0$, factoring out the nonzero exponential term $e^{rt}$, and setting the remaining polynomial in $r$ to zero.

Conditions: The differential equation is $ay'' + by' + cy = 0$.; The trial solution is $y = e^{rt}$.; $a, b, c$ are constants.

The characteristic equation is the quadratic algebraic equation $0 = ar^2 + br + c$. It is obtained by substituting the trial solution $y = e^{rt}$ into the second-order linear homogeneous differential equation $ay'' + by' + cy = 0$, factoring out the nonzero exponential term $e^{rt}$, and setting the remaining polynomial in $r$ to zero.

Conditions: The differential equation is $ay'' + by' + cy = 0$.; The trial solution is $y = e^{rt}$.; $a, b, c$ are constants.

The characteristic equation $ar^2 + br + c = 0$ is solved by applying the quadratic formula. This yields the roots $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, which provide the values for the parameter $r$ in the exponential trial solution.

Conditions: The characteristic equation is quadratic in $r$.; $a \neq 0$.

The characteristic equation $ar^2 + br + c = 0$ is solved by applying the quadratic formula. This yields the roots $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, which provide the values for the parameter $r$ in the exponential trial solution.

Conditions: The characteristic equation is quadratic in $r$.; $a \neq 0$.