(7r+8)(r-8)=0

Solution for (7r+8)(r-8)=0 equation:

(7r+8)(r-8)=0

We multiply parentheses ..

(+7r^2-56r+8r-64)=0

We get rid of parentheses

7r^2-56r+8r-64=0

We add all the numbers together, and all the variables

7r^2-48r-64=0

a = 7; b = -48; c = -64;
Δ = b2-4ac
Δ = -482-4·7·(-64)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-64}{2*7}=\frac{-16}{14}
=-1+1/7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+64}{2*7}=\frac{112}{14}
=8
$