+0  
 
0
139
3
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says there could be two solutions, round four decimal places.  Thanks

 Mar 2, 2022
 #3
avatar+117834 
+1

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,

 Mar 2, 2022

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