81.3k views
2 votes
Find the zeros of the following quadratic function y = -6x^2 + 100x - 180

User Abrar
by
8.4k points

1 Answer

3 votes
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

In this equation, \(a = -6\), \(b = 100\), and \(c = -180\).

Now, plug these values into the quadratic formula:

\[x = \frac{-100 \pm \sqrt{100^2 - 4(-6)(-180)}}{2(-6)}\]

Simplify the equation to find the zeros:

\[x = \frac{-100 \pm \sqrt{10000 - 4320}}{-12}\]

\[x = \frac{-100 \pm \sqrt{5680}}{-12}\]

Now, you can simplify further:

\[x = \frac{-100 \pm 2\sqrt{1420}}{-12}\]

\[x = \frac{-50 \pm \sqrt{355}}{-6}\]

So, the zeros of the quadratic function are:

\[x_1 = \frac{-50 + \sqrt{355}}{-6}\]

\[x_2 = \frac{-50 - \sqrt{355}}{-6}\]
User Cpp Forever
by
7.6k points

No related questions found