3y^2+7=19

Solution for 3y^2+7=19 equation:

3y^2+7=19

We move all terms to the left:

3y^2+7-(19)=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
$