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

Solution for (6+r)(10+3r)=0 equation:

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

We add all the numbers together, and all the variables

(r+6)(3r+10)=0

We multiply parentheses ..

(+3r^2+10r+18r+60)=0

We get rid of parentheses

3r^2+10r+18r+60=0

We add all the numbers together, and all the variables

3r^2+28r+60=0

a = 3; b = 28; c = +60;
Δ = b2-4ac
Δ = 282-4·3·60
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-8}{2*3}=\frac{-36}{6}
=-6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+8}{2*3}=\frac{-20}{6}
=-3+1/3
$