(7r+1)(3r+4)=0

Solution for (7r+1)(3r+4)=0 equation:

(7r+1)(3r+4)=0

We multiply parentheses ..

(+21r^2+28r+3r+4)=0

We get rid of parentheses

21r^2+28r+3r+4=0

We add all the numbers together, and all the variables

21r^2+31r+4=0

a = 21; b = 31; c = +4;
Δ = b2-4ac
Δ = 312-4·21·4
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-25}{2*21}=\frac{-56}{42}
=-1+1/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+25}{2*21}=\frac{-6}{42}
=-1/7
$