How to find zeros for y=8x^2+12x+2

OK, Stacy, we have

8x^2  + 12x +2 = 0   Let's divide by 2 to reduce the coefficients

4x^2 + 6x + 1 = 0

This won't factor, so let's put it into the on-site solver...BTW...I know it has "real" solutions....

$${\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{1.309\: \!016\: \!994\: \!374\: \!947\: \!4}}\\
{\mathtt{x}} = -{\mathtt{0.190\: \!983\: \!005\: \!625\: \!052\: \!6}}\\
\end{array} \right\}$$And there are your two answers using the quadratic formula.... 

Hi Stacy

You need to use the quadratic formula.

https://www.youtube.com/watch?v=O8ezDEk3qCg

y=8x^2+12x+2

a=8

b=12

c=2

Just sub the values in and calculate the answers

CPhill has already told you what those answers should be.