How do you find the zeros, real and imaginary, of y= 3x^2+18x-24 using the quadratic formula?

Feb 17, 2016

In this case both roots are real:

$x = \frac{- 18 - 24.7}{6} \approx - 7.1$ and $x = \frac{- 18 + 24.7}{6} \approx 1.1$

Explanation:

The zeroes, or solutions, are the points where the line crosses the $x$ axis, the line $y = 0$.

Set the equation equal to $0$:

$3 {x}^{2} + 18 x - 24 = 0$

This is the standard form of a quadratic equation:

$a {x}^{2} + b x + c = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$= \frac{- 18 \pm \sqrt{{18}^{2} - 4 \cdot 3 \cdot \left(- 24\right)}}{2 \cdot 3}$
Notice the piece under the square root sign (radical). If it was $0$ there would only be one root, if it was negative the roots would be imaginary, but since it is positive both the roots are real.
$x = \frac{- 18 \pm \sqrt{612}}{6}$
$x = \frac{- 18 - 24.7}{6} \approx - 7.1$ and $x = \frac{- 18 + 24.7}{6} \approx 1.1$