-2r^2-12r+22=0

Solution for -2r^2-12r+22=0 equation:

-2r^2-12r+22=0

a = -2; b = -12; c = +22;
Δ = b2-4ac
Δ = -122-4·(-2)·22
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{5}}{2*-2}=\frac{12-8\sqrt{5}}{-4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{5}}{2*-2}=\frac{12+8\sqrt{5}}{-4}
$