(3r-7)(5r+1)=0

Solution for (3r-7)(5r+1)=0 equation:

(3r-7)(5r+1)=0

We multiply parentheses ..

(+15r^2+3r-35r-7)=0

We get rid of parentheses

15r^2+3r-35r-7=0

We add all the numbers together, and all the variables

15r^2-32r-7=0

a = 15; b = -32; c = -7;
Δ = b2-4ac
Δ = -322-4·15·(-7)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-38}{2*15}=\frac{-6}{30}
=-1/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+38}{2*15}=\frac{70}{30}
=2+1/3
$