says there could be two solutions, round four decimal places. Thanks
g(x) = -.03x^2+x+6=0
0= -.03x^2+x+6
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-1 \pm \sqrt{1^2-4*-0.03*6} \over 2*-0.03}\\ x = {-1 \pm \sqrt{1+0.72} \over -0.06}\\ \)
you can simplify it,
