(2y+5)(y+4)=2

Solution for (2y+5)(y+4)=2 equation:

(2y+5)(y+4)=2

We move all terms to the left:

(2y+5)(y+4)-(2)=0

We multiply parentheses ..

(+2y^2+8y+5y+20)-2=0

We get rid of parentheses

2y^2+8y+5y+20-2=0

We add all the numbers together, and all the variables

2y^2+13y+18=0

a = 2; b = 13; c = +18;
Δ = b2-4ac
Δ = 132-4·2·18
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*2}=\frac{-18}{4}
=-4+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*2}=\frac{-8}{4}
=-2
$