(5r+3)(r+10)=0

Solution for (5r+3)(r+10)=0 equation:

(5r+3)(r+10)=0

We multiply parentheses ..

(+5r^2+50r+3r+30)=0

We get rid of parentheses

5r^2+50r+3r+30=0

We add all the numbers together, and all the variables

5r^2+53r+30=0

a = 5; b = 53; c = +30;
Δ = b2-4ac
Δ = 532-4·5·30
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(53)-47}{2*5}=\frac{-100}{10}
=-10
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(53)+47}{2*5}=\frac{-6}{10}
=-3/5
$