(3r-5)(2r+1)=0

Solution for (3r-5)(2r+1)=0 equation:

(3r-5)(2r+1)=0

We multiply parentheses ..

(+6r^2+3r-10r-5)=0

We get rid of parentheses

6r^2+3r-10r-5=0

We add all the numbers together, and all the variables

6r^2-7r-5=0

a = 6; b = -7; c = -5;
Δ = b2-4ac
Δ = -72-4·6·(-5)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-13}{2*6}=\frac{-6}{12}
=-1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+13}{2*6}=\frac{20}{12}
=1+2/3
$