#### Answer

$y=5,-1$

#### Work Step by Step

$2y^2-8y-10$
Rewrite the equation as:
$y^2-4y=5$
Compare it with the standard form of quadratic equation $ax^2+bx+c$, we have $a=1, b=-4$
Therefore, $b^2=4ac$ $\implies$ $c=\dfrac{b^2}{4a}$
Thus, $c=\dfrac{b^2}{4a}=\dfrac{(-4)^2}{4}=4$
To complete the square, add $1$ on both sides.
$y^2-4y+4=5+4$
$\implies (y-2)^2=9$
$\implies y-2=3$
and
$\implies y-2=-3$
or, $y=5,-1$