(3+y)(5y+2)=0

Solution for (3+y)(5y+2)=0 equation:

(3+y)(5y+2)=0

We add all the numbers together, and all the variables

(y+3)(5y+2)=0

We multiply parentheses ..

(+5y^2+2y+15y+6)=0

We get rid of parentheses

5y^2+2y+15y+6=0

We add all the numbers together, and all the variables

5y^2+17y+6=0

a = 5; b = 17; c = +6;
Δ = b2-4ac
Δ = 172-4·5·6
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-13}{2*5}=\frac{-30}{10}
=-3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+13}{2*5}=\frac{-4}{10}
=-2/5
$