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

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

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

We multiply parentheses ..

(+12r^2+21r+4r+7)=0

We get rid of parentheses

12r^2+21r+4r+7=0

We add all the numbers together, and all the variables

12r^2+25r+7=0

a = 12; b = 25; c = +7;
Δ = b2-4ac
Δ = 252-4·12·7
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{-(25)-17}{2*12}=\frac{-42}{24}
=-1+3/4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+17}{2*12}=\frac{-8}{24}
=-1/3
$