2y^2+1=129

Solution for 2y^2+1=129 equation:

2y^2+1=129

We move all terms to the left:

2y^2+1-(129)=0

We add all the numbers together, and all the variables

2y^2-128=0

a = 2; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·2·(-128)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*2}=\frac{-32}{4}
=-8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*2}=\frac{32}{4}
=8
$