(r-6)(r+1)=0

Solution for (r-6)(r+1)=0 equation:

(r-6)(r+1)=0

We multiply parentheses ..

(+r^2+r-6r-6)=0

We get rid of parentheses

r^2+r-6r-6=0

We add all the numbers together, and all the variables

r^2-5r-6=0

a = 1; b = -5; c = -6;
Δ = b2-4ac
Δ = -52-4·1·(-6)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*1}=\frac{-2}{2}
=-1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*1}=\frac{12}{2}
=6
$