(13r+8)(5r-3)=0

Solution for (13r+8)(5r-3)=0 equation:

(13r+8)(5r-3)=0

We multiply parentheses ..

(+65r^2-39r+40r-24)=0

We get rid of parentheses

65r^2-39r+40r-24=0

We add all the numbers together, and all the variables

65r^2+r-24=0

a = 65; b = 1; c = -24;
Δ = b2-4ac
Δ = 12-4·65·(-24)
Δ = 6241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6241}=79$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-79}{2*65}=\frac{-80}{130}
=-8/13
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+79}{2*65}=\frac{78}{130}
=3/5
$