I need help finding the Xs for this problem and thank u

Solution
For this case we have the following equation:
[tex]y=3x^2+16x-12[/tex]a= 3, b= 16, c= -12
Then we can use the quadratic formula given by:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Replacing we got:
[tex]x=\frac{-16\pm\sqrt[]{16^2-4(3\cdot-12)}}{2\cdot3}=\frac{-16\pm\sqrt[]{400}}{6}=\frac{-16\pm20}{6}[/tex]Then the final answer would be:
x= 4/6= 2/3
x= -6