First, let's divide the equation by 6:
[tex]\begin{gathered} 6x^2+7x-20=0 \\ \rightarrow x^2+\frac{7}{6}x-\frac{10}{3}=0 \end{gathered}[/tex]
Completing the square,
[tex]\begin{gathered} x^2+\frac{7}{6}x-\frac{10}{3}=0 \\ \\ \rightarrow x^2+\frac{7}{6}x-\frac{10}{3}+\frac{529}{144}=\frac{529}{144} \\ \\ \rightarrow x^2+\frac{7}{6}x+\frac{49}{144}=\frac{529}{144} \\ \\ \rightarrow(x+\frac{7}{12})^2=\frac{529}{144} \end{gathered}[/tex]
To input the expression, we would have:
[tex](x+0.58)^2=3.67[/tex]
Solving for x,
[tex]\begin{gathered} (x+\frac{7}{12})^2=\frac{529}{144} \\ \\ \rightarrow x+\frac{7}{12}=\pm\text{ }\sqrt[]{\frac{529}{144}} \\ \\ \rightarrow x=-\frac{7}{12}\pm\frac{23}{12} \\ \\ \rightarrow x_1=\frac{4}{3} \\ \rightarrow x_2=-\frac{5}{2} \end{gathered}[/tex]
To input this answer, we'll have:
x = -2.5 and x = 1.33