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

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

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

We multiply parentheses ..

(+8r^2-3r+64r-24)=0

We get rid of parentheses

8r^2-3r+64r-24=0

We add all the numbers together, and all the variables

8r^2+61r-24=0

a = 8; b = 61; c = -24;
Δ = b2-4ac
Δ = 612-4·8·(-24)
Δ = 4489
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4489}=67$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(61)-67}{2*8}=\frac{-128}{16}
=-8
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(61)+67}{2*8}=\frac{6}{16}
=3/8
$