(11r+7)(4r-12)=0

Solution for (11r+7)(4r-12)=0 equation:

(11r+7)(4r-12)=0

We multiply parentheses ..

(+44r^2-132r+28r-84)=0

We get rid of parentheses

44r^2-132r+28r-84=0

We add all the numbers together, and all the variables

44r^2-104r-84=0

a = 44; b = -104; c = -84;
Δ = b2-4ac
Δ = -1042-4·44·(-84)
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-104)-160}{2*44}=\frac{-56}{88}
=-7/11
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-104)+160}{2*44}=\frac{264}{88}
=3
$