11r^2+3r-18=0

Solution for 11r^2+3r-18=0 equation:

11r^2+3r-18=0

a = 11; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·11·(-18)
Δ = 801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{801}=\sqrt{9*89}=\sqrt{9}*\sqrt{89}=3\sqrt{89}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{89}}{2*11}=\frac{-3-3\sqrt{89}}{22}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{89}}{2*11}=\frac{-3+3\sqrt{89}}{22}
$