(r+6)(5r-4)=0

Solution for (r+6)(5r-4)=0 equation:

(r+6)(5r-4)=0

We multiply parentheses ..

(+5r^2-4r+30r-24)=0

We get rid of parentheses

5r^2-4r+30r-24=0

We add all the numbers together, and all the variables

5r^2+26r-24=0

a = 5; b = 26; c = -24;
Δ = b2-4ac
Δ = 262-4·5·(-24)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-34}{2*5}=\frac{-60}{10}
=-6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+34}{2*5}=\frac{8}{10}
=4/5
$