Answer: 0
Step-by-step explanation:
The given equation: [tex]6a^2+5a+4=3[/tex]
Subtract 3 from both the sides, we get
[tex]6a^2+5a+1=0[/tex]
Now , we can split 5a as 2a+3a and [tex]2a\times 3a = 6a^2[/tex]
So, [tex]6a^2+5a+1=0\Rightarrow\ 6a^2+2a+3a+1=0[/tex]
[tex]\Rightarrow\ 2a(3a+1)+(3a+1)=0\\\\\Rightarrow\ (3a+1)(2a+1)=0\\\\\Rightarrow\ (3a+1)=0\text{ or }(2a+1)=0\\\\\Rightarrow\ a=-\dfrac{1}{3}\text{ or }a=-\dfrac{1}{2}[/tex]
At [tex]a=-\dfrac{1}{3}[/tex]
[tex]2a+1=2(-\dfrac{1}{3})+1=-\dfrac{2}{3}+1=\dfrac{-2+3}{3}=\dfrac{1}3{}[/tex]
At [tex]a=-\dfrac{1}{2}[/tex]
[tex]2a+1=2(-\dfrac{1}{2})+1=-1+1=0[/tex]
Since, [tex]0< \dfrac{1}{3}[/tex]
Hence, the possible value of 2a+1 is 0.
