we have the equation
6a²=5+a
group terms
6a^2-a-5=0
solve using the quadratic formula
we have
a=6
b=-1
c=-5
substitute
[tex]a=\frac{-(-1)\pm\sqrt[]{-1^2-4(6)(-5)}}{2(6)}[/tex][tex]\begin{gathered} a=\frac{1\pm\sqrt[]{121}}{12} \\ \\ a=\frac{1\pm11}{12} \end{gathered}[/tex]the solutions are
a=1 and
a=-10/12=-5/6
therefore
answer is
