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

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

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

We multiply parentheses ..

(+10r^2-2r+15r-3)=0

We get rid of parentheses

10r^2-2r+15r-3=0

We add all the numbers together, and all the variables

10r^2+13r-3=0

a = 10; b = 13; c = -3;
Δ = b2-4ac
Δ = 132-4·10·(-3)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*10}=\frac{-30}{20}
=-1+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*10}=\frac{4}{20}
=1/5
$