103=3y^2-5

Solution for 103=3y^2-5 equation:

103=3y^2-5

We move all terms to the left:

103-(3y^2-5)=0

We get rid of parentheses

-3y^2+5+103=0

We add all the numbers together, and all the variables

-3y^2+108=0

a = -3; b = 0; c = +108;
Δ = b2-4ac
Δ = 02-4·(-3)·108
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36}{2*-3}=\frac{-36}{-6}
=+6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36}{2*-3}=\frac{36}{-6}
=-6
$