(y+1)(2y+2)=29

Solution for (y+1)(2y+2)=29 equation:

(y+1)(2y+2)=29

We move all terms to the left:

(y+1)(2y+2)-(29)=0

We multiply parentheses ..

(+2y^2+2y+2y+2)-29=0

We get rid of parentheses

2y^2+2y+2y+2-29=0

We add all the numbers together, and all the variables

2y^2+4y-27=0

a = 2; b = 4; c = -27;
Δ = b2-4ac
Δ = 42-4·2·(-27)
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{58}}{2*2}=\frac{-4-2\sqrt{58}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{58}}{2*2}=\frac{-4+2\sqrt{58}}{4}
$