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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2r^2+11r+15=0

a = 2; b = 11; c = +15;
Δ = b2-4ac
Δ = 112-4·2·15
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-1}{2*2}=\frac{-12}{4}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*2}=\frac{-10}{4}
=-2+1/2
$