Question

I need help finding the Xs on this question....thank you

Answer 1

In order to find the zeros of this quadratic function, we can use the quadratic formula:

`\begin{gathered} x_1=\frac{-b+\sqrt[]{b^2-4ac}}{2a} \\ x_2=\frac{-b-\sqrt[]{b^2-4ac}}{2a} \end{gathered}`

Using the given values of a = 3, b = 16 and c = -12, we have:

`\begin{gathered} x_1=\frac{-16+\sqrt[]{256+144}}{2\cdot3}=(-16+20)/(6)=(2)/(3) \\ x_2=(-16-20)/(6)=-6 \end{gathered}`

So the values of x that make this equation be equal zero are x = 2/3 and x = -6.