(r+5)(8r-5)=0

Solution for (r+5)(8r-5)=0 equation:

(r+5)(8r-5)=0

We multiply parentheses ..

(+8r^2-5r+40r-25)=0

We get rid of parentheses

8r^2-5r+40r-25=0

We add all the numbers together, and all the variables

8r^2+35r-25=0

a = 8; b = 35; c = -25;
Δ = b2-4ac
Δ = 352-4·8·(-25)
Δ = 2025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2025}=45$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-45}{2*8}=\frac{-80}{16}
=-5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+45}{2*8}=\frac{10}{16}
=5/8
$