-19=-10+6r+13r+10r^2

Solution for -19=-10+6r+13r+10r^2 equation:

-19=-10+6r+13r+10r^2

We move all terms to the left:

-19-(-10+6r+13r+10r^2)=0

We get rid of parentheses

-10r^2-6r-13r+10-19=0

We add all the numbers together, and all the variables

-10r^2-19r-9=0

a = -10; b = -19; c = -9;
Δ = b2-4ac
Δ = -192-4·(-10)·(-9)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-1}{2*-10}=\frac{18}{-20}
=-9/10
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+1}{2*-10}=\frac{20}{-20}
=-1
$