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

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

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

We multiply parentheses ..

(+5r^2+20r+3r+12)=0

We get rid of parentheses

5r^2+20r+3r+12=0

We add all the numbers together, and all the variables

5r^2+23r+12=0

a = 5; b = 23; c = +12;
Δ = b2-4ac
Δ = 232-4·5·12
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-17}{2*5}=\frac{-40}{10}
=-4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*5}=\frac{-6}{10}
=-3/5
$