(2y)^2+y^2=25

Solution for (2y)^2+y^2=25 equation:

(2y)^2+y^2=25

We move all terms to the left:

(2y)^2+y^2-(25)=0

We add all the numbers together, and all the variables

3y^2-25=0

a = 3; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·3·(-25)
Δ = 300
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:

$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{3}}{2*3}=\frac{0-10\sqrt{3}}{6}
=-\frac{10\sqrt{3}}{6}
=-\frac{5\sqrt{3}}{3}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{3}}{2*3}=\frac{0+10\sqrt{3}}{6}
=\frac{10\sqrt{3}}{6}
=\frac{5\sqrt{3}}{3}
$