+13 \(y= -6x^2+42x -6\)
First, factor out a -6
\(y = -6(x^2 - 7x + 1)\)
Then, set the value in the parentheses equal to 0
\(x^2-7x+1=0\)
Now, use the quadratic formula to solve
\(y = {(7) \pm \sqrt{(7)^2-4(1)(1)} \over 2(1)}\)
\(y = {7 \pm 3 \sqrt5 \over 2}\)
\(y \approx 6.854, 0.146\)
Your zeros are approximately 6.854, and 0.146
