2y2+10=210

Solution for 2y2+10=210 equation:

2y^2+10=210

We move all terms to the left:

2y^2+10-(210)=0

We add all the numbers together, and all the variables

2y^2-200=0

a = 2; b = 0; c = -200;
Δ = b2-4ac
Δ = 02-4·2·(-200)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*2}=\frac{-40}{4}
=-10
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*2}=\frac{40}{4}
=10
$