(7r+1)(r+7)=0

Solution for (7r+1)(r+7)=0 equation:

(7r+1)(r+7)=0

We multiply parentheses ..

(+7r^2+49r+r+7)=0

We get rid of parentheses

7r^2+49r+r+7=0

We add all the numbers together, and all the variables

7r^2+50r+7=0

a = 7; b = 50; c = +7;
Δ = b2-4ac
Δ = 502-4·7·7
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-48}{2*7}=\frac{-98}{14}
=-7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+48}{2*7}=\frac{-2}{14}
=-1/7
$