3y2+7=199

Solution for 3y2+7=199 equation:

3y^2+7=199

We move all terms to the left:

3y^2+7-(199)=0

We add all the numbers together, and all the variables

3y^2-192=0

a = 3; b = 0; c = -192;
Δ = b2-4ac
Δ = 02-4·3·(-192)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*3}=\frac{-48}{6}
=-8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*3}=\frac{48}{6}
=8
$