65=(8y)^2+y^2

Solution for 65=(8y)^2+y^2 equation:

65=(8y)^2+y^2

We move all terms to the left:

65-((8y)^2+y^2)=0

We get rid of parentheses

-8y^2-y^2+65=0

We add all the numbers together, and all the variables

-9y^2+65=0

a = -9; b = 0; c = +65;
Δ = b2-4ac
Δ = 02-4·(-9)·65
Δ = 2340
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2340}=\sqrt{36*65}=\sqrt{36}*\sqrt{65}=6\sqrt{65}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{65}}{2*-9}=\frac{0-6\sqrt{65}}{-18}
=-\frac{6\sqrt{65}}{-18}
=-\frac{\sqrt{65}}{-3}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{65}}{2*-9}=\frac{0+6\sqrt{65}}{-18}
=\frac{6\sqrt{65}}{-18}
=\frac{\sqrt{65}}{-3}
$