How to find all the real solutions of the polynomial equation 2 y 4 +...

${\displaystyle 2y^4+7y^3-26y^2+23y-6}$
Keep in mind that ${\displaystyle 0}$ is the total of the coefficients. Which is:
${\displaystyle 2+7-26+23-6=0}$
So ${\displaystyle y=1}$ is a zero and ${\displaystyle (y-1)}$ a factor:
${\displaystyle 2y^4+7y^3-26y^2+23y-6=(y-1)(2y^3+9y^2-17y+6)}$
The sum of the coefficients of the remaining cubic factor also sum to ${\displaystyle 0}$. That is:
${\displaystyle 2+9-17+6=0}$
So ${\displaystyle y=1}$ is a zero again and ${\displaystyle (y-1)}$ a factor:
${\displaystyle 2y^3+9y^2-17y+6=(y-1)(2y^2+11y-6)}$
We can factor the remaining quadratic using an AC method:
Find a pair of factors of ${\displaystyle AC=2\cdot 6=12}$ which differ by ${\displaystyle B=11}$
The pair ${\displaystyle 12,1}$ works.
Split the middle term using that pair, then factor by grouping:
${\displaystyle 2y^2+11y-6}$
${\displaystyle =2y^2+12y-y-6}$
${\displaystyle =(2y^2+12y)-(y+6)}$
${\displaystyle =2y(y+6)-1(y+6)}$
${\displaystyle =(2y-1)(y+6)}$
So the remaining zeros are ${\displaystyle y=\frac{1}{2}}$ and ${\displaystyle y=-6}$