11r^2+14r-2685=0

Solution for 11r^2+14r-2685=0 equation:

11r^2+14r-2685=0

a = 11; b = 14; c = -2685;
Δ = b2-4ac
Δ = 142-4·11·(-2685)
Δ = 118336
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{118336}=344$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-344}{2*11}=\frac{-358}{22}
=-16+3/11
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+344}{2*11}=\frac{330}{22}
=15
$