(y+1)(y+2)=5

Solution for (y+1)(y+2)=5 equation:

(y+1)(y+2)=5

We move all terms to the left:

(y+1)(y+2)-(5)=0

We multiply parentheses ..

(+y^2+2y+y+2)-5=0

We get rid of parentheses

y^2+2y+y+2-5=0

We add all the numbers together, and all the variables

y^2+3y-3=0

a = 1; b = 3; c = -3;
Δ = b2-4ac
Δ = 32-4·1·(-3)
Δ = 21
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{21}}{2*1}=\frac{-3-\sqrt{21}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{21}}{2*1}=\frac{-3+\sqrt{21}}{2}
$