11-19r-6r^2=0

Solution for 11-19r-6r^2=0 equation:

11-19r-6r^2=0

a = -6; b = -19; c = +11;
Δ = b2-4ac
Δ = -192-4·(-6)·11
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-25}{2*-6}=\frac{-6}{-12}
=1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+25}{2*-6}=\frac{44}{-12}
=-3+2/3
$