Any quadratic equations can also be solved with the quadratic formula II.

i

### Hint

In contrast to the quadratic formula I the quadratic equations in the quadratic formula IIdo not need to be in the canonical form.

Depending on the federal state, it is different whether the quadratic formula I or the quadratic formula II is taught. Ultimately, however, you get the same result with both formulas.

Given is a quadratic equation in the form: $\color{red}{a}x^2+\color{green}{b}x+\color{blue}{c}=0$.
The quadratic formula II for solving this equation is:

$x_{1,2} = \frac{-\color{green}{b} \pm \sqrt{\color{green}{b}^2 - 4\color{red}{a}\color{blue}{c}}}{2\color{red}{a}}$

### Example

Solve quadratic equation: $\color{red}{3}x^2+\color{green}{18}x+\color{blue}{15}=0$

1. #### Insert $a$, $b$ and $c$ in the quadratic formula II

$x_{1,2} = \frac{-\color{green}{b} \pm \sqrt{\color{green}{b}^2 - 4\color{red}{a}\color{blue}{c}}}{2\color{red}{a}}$
$x_{1,2} = \frac{-\color{green}{18} \pm \sqrt{\color{green}{18}^2 - 4\cdot\color{red}{3}\cdot\color{blue}{15}}}{2\cdot\color{red}{3}}$
2. #### Simplify term

$x_{1,2} = \frac{-18 \pm \sqrt{324 - 180}}{6}$
$x_{1,2} = \frac{-18 \pm \sqrt{144}}{6}$
$x_{1,2} = \frac{-18 \pm 12}{6}$
3. #### Calculate solutions

$x_{1} = \frac{-18 + 12}{6} = \frac{-6}{6}=-1$
$x_{2} = \frac{-18 - 12}{6} = \frac{-30}{6}=-5$