(y+1)(3y-1)=666

Solution for (y+1)(3y-1)=666 equation:

(y+1)(3y-1)=666

We move all terms to the left:

(y+1)(3y-1)-(666)=0

We multiply parentheses ..

(+3y^2-1y+3y-1)-666=0

We get rid of parentheses

3y^2-1y+3y-1-666=0

We add all the numbers together, and all the variables

3y^2+2y-667=0

a = 3; b = 2; c = -667;
Δ = b2-4ac
Δ = 22-4·3·(-667)
Δ = 8008
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:

$\sqrt{\Delta}=\sqrt{8008}=\sqrt{4*2002}=\sqrt{4}*\sqrt{2002}=2\sqrt{2002}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{2002}}{2*3}=\frac{-2-2\sqrt{2002}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{2002}}{2*3}=\frac{-2+2\sqrt{2002}}{6}
$