r(7r+6)=0

Solution for r(7r+6)=0 equation:

r(7r+6)=0

We multiply parentheses

7r^2+6r=0

a = 7; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·7·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*7}=\frac{-12}{14}
=-6/7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*7}=\frac{0}{14}
=0
$