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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5r^2-7r-6=0

a = 5; b = -7; c = -6;
Δ = b2-4ac
Δ = -72-4·5·(-6)
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*5}=\frac{-6}{10}
=-3/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+13}{2*5}=\frac{20}{10}
=2
$