r^2-12r+22=0

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

r^2-12r+22=0

a = 1; b = -12; c = +22;
Δ = b2-4ac
Δ = -122-4·1·22
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{14}}{2*1}=\frac{12-2\sqrt{14}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{14}}{2*1}=\frac{12+2\sqrt{14}}{2}
$