r^2-10r+22=6

Solution for r^2-10r+22=6 equation:

r^2-10r+22=6

We move all terms to the left:

r^2-10r+22-(6)=0

We add all the numbers together, and all the variables

r^2-10r+16=0

a = 1; b = -10; c = +16;
Δ = b2-4ac
Δ = -102-4·1·16
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6}{2*1}=\frac{4}{2}
=2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6}{2*1}=\frac{16}{2}
=8
$