2r^2+11r-6=0

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

2r^2+11r-6=0

a = 2; b = 11; c = -6;
Δ = b2-4ac
Δ = 112-4·2·(-6)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-13}{2*2}=\frac{-24}{4}
=-6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*2}=\frac{2}{4}
=1/2
$