(4r-8)(3r+9)=0

Solution for (4r-8)(3r+9)=0 equation:

(4r-8)(3r+9)=0

We multiply parentheses ..

(+12r^2+36r-24r-72)=0

We get rid of parentheses

12r^2+36r-24r-72=0

We add all the numbers together, and all the variables

12r^2+12r-72=0

a = 12; b = 12; c = -72;
Δ = b2-4ac
Δ = 122-4·12·(-72)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-60}{2*12}=\frac{-72}{24}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+60}{2*12}=\frac{48}{24}
=2
$