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

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

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

We multiply parentheses ..

(+4y^2+10y+6y+15)=0

We get rid of parentheses

4y^2+10y+6y+15=0

We add all the numbers together, and all the variables

4y^2+16y+15=0

a = 4; b = 16; c = +15;
Δ = b2-4ac
Δ = 162-4·4·15
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4}{2*4}=\frac{-20}{8}
=-2+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4}{2*4}=\frac{-12}{8}
=-1+1/2
$