(8r+3)(6r+5)=0

Solution for (8r+3)(6r+5)=0 equation:

(8r+3)(6r+5)=0

We multiply parentheses ..

(+48r^2+40r+18r+15)=0

We get rid of parentheses

48r^2+40r+18r+15=0

We add all the numbers together, and all the variables

48r^2+58r+15=0

a = 48; b = 58; c = +15;
Δ = b2-4ac
Δ = 582-4·48·15
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(58)-22}{2*48}=\frac{-80}{96}
=-5/6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(58)+22}{2*48}=\frac{-36}{96}
=-3/8
$