(3y)^2+y^2=10

Solution for (3y)^2+y^2=10 equation:

(3y)^2+y^2=10

We move all terms to the left:

(3y)^2+y^2-(10)=0

We add all the numbers together, and all the variables

4y^2-10=0

a = 4; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·4·(-10)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*4}=\frac{0-4\sqrt{10}}{8}
=-\frac{4\sqrt{10}}{8}
=-\frac{\sqrt{10}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*4}=\frac{0+4\sqrt{10}}{8}
=\frac{4\sqrt{10}}{8}
=\frac{\sqrt{10}}{2}
$