3(y^2+3)=28

Solution for 3(y^2+3)=28 equation:

3(y^2+3)=28

We move all terms to the left:

3(y^2+3)-(28)=0

We multiply parentheses

3y^2+9-28=0

We add all the numbers together, and all the variables

3y^2-19=0

a = 3; b = 0; c = -19;
Δ = b2-4ac
Δ = 02-4·3·(-19)
Δ = 228
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{57}}{2*3}=\frac{0-2\sqrt{57}}{6}
=-\frac{2\sqrt{57}}{6}
=-\frac{\sqrt{57}}{3}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{57}}{2*3}=\frac{0+2\sqrt{57}}{6}
=\frac{2\sqrt{57}}{6}
=\frac{\sqrt{57}}{3}
$