11r^2+3r-27=0

Solution for 11r^2+3r-27=0 equation:

11r^2+3r-27=0

a = 11; b = 3; c = -27;
Δ = b2-4ac
Δ = 32-4·11·(-27)
Δ = 1197
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:

$\sqrt{\Delta}=\sqrt{1197}=\sqrt{9*133}=\sqrt{9}*\sqrt{133}=3\sqrt{133}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{133}}{2*11}=\frac{-3-3\sqrt{133}}{22}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{133}}{2*11}=\frac{-3+3\sqrt{133}}{22}
$