0=(3+y)(5y+4)

Solution for 0=(3+y)(5y+4) equation:

0=(3+y)(5y+4)

We move all terms to the left:

0-((3+y)(5y+4))=0

We add all the numbers together, and all the variables

-((y+3)(5y+4))+0=0

We add all the numbers together, and all the variables

-((y+3)(5y+4))=0

We multiply parentheses ..

-((+5y^2+4y+15y+12))=0


We calculate terms in parentheses: -((+5y^2+4y+15y+12)), so:

(+5y^2+4y+15y+12)

We get rid of parentheses

5y^2+4y+15y+12

We add all the numbers together, and all the variables

5y^2+19y+12

Back to the equation:

-(5y^2+19y+12)


We get rid of parentheses

-5y^2-19y-12=0

a = -5; b = -19; c = -12;
Δ = b2-4ac
Δ = -192-4·(-5)·(-12)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*-5}=\frac{8}{-10}
=-4/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*-5}=\frac{30}{-10}
=-3
$