(7r+5)(2r-7)=0

Solution for (7r+5)(2r-7)=0 equation:

(7r+5)(2r-7)=0

We multiply parentheses ..

(+14r^2-49r+10r-35)=0

We get rid of parentheses

14r^2-49r+10r-35=0

We add all the numbers together, and all the variables

14r^2-39r-35=0

a = 14; b = -39; c = -35;
Δ = b2-4ac
Δ = -392-4·14·(-35)
Δ = 3481
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3481}=59$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-59}{2*14}=\frac{-20}{28}
=-5/7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+59}{2*14}=\frac{98}{28}
=3+1/2
$