2+4y^2=y^2+14

Solution for 2+4y^2=y^2+14 equation:

2+4y^2=y^2+14

We move all terms to the left:

2+4y^2-(y^2+14)=0

We get rid of parentheses

4y^2-y^2-14+2=0

We add all the numbers together, and all the variables

3y^2-12=0

a = 3; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·3·(-12)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*3}=\frac{-12}{6}
=-2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*3}=\frac{12}{6}
=2
$