2y^2+y^2=18

Solution for 2y^2+y^2=18 equation:

2y^2+y^2=18

We move all terms to the left:

2y^2+y^2-(18)=0

We add all the numbers together, and all the variables

3y^2-18=0

a = 3; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·3·(-18)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{6}}{2*3}=\frac{0-6\sqrt{6}}{6}
=-\frac{6\sqrt{6}}{6}
=-\sqrt{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{6}}{2*3}=\frac{0+6\sqrt{6}}{6}
=\frac{6\sqrt{6}}{6}
=\sqrt{6}
$