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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

9y^2-28y+9=0

a = 9; b = -28; c = +9;
Δ = b2-4ac
Δ = -282-4·9·9
Δ = 460
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:

$\sqrt{\Delta}=\sqrt{460}=\sqrt{4*115}=\sqrt{4}*\sqrt{115}=2\sqrt{115}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-2\sqrt{115}}{2*9}=\frac{28-2\sqrt{115}}{18}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+2\sqrt{115}}{2*9}=\frac{28+2\sqrt{115}}{18}
$