5r(3r-7)=0

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

5r(3r-7)=0

We multiply parentheses

15r^2-35r=0

a = 15; b = -35; c = 0;
Δ = b2-4ac
Δ = -352-4·15·0
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-35}{2*15}=\frac{0}{30}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+35}{2*15}=\frac{70}{30}
=2+1/3
$