(5r+6)(r-8)=0

Solution for (5r+6)(r-8)=0 equation:

(5r+6)(r-8)=0

We multiply parentheses ..

(+5r^2-40r+6r-48)=0

We get rid of parentheses

5r^2-40r+6r-48=0

We add all the numbers together, and all the variables

5r^2-34r-48=0

a = 5; b = -34; c = -48;
Δ = b2-4ac
Δ = -342-4·5·(-48)
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-46}{2*5}=\frac{-12}{10}
=-1+1/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+46}{2*5}=\frac{80}{10}
=8
$