r2-10r+12=-10

Solution for r2-10r+12=-10 equation:

r2-10r+12=-10

We move all terms to the left:

r2-10r+12-(-10)=0

We add all the numbers together, and all the variables

r^2-10r+22=0

a = 1; b = -10; c = +22;
Δ = b2-4ac
Δ = -102-4·1·22
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{3}}{2*1}=\frac{10-2\sqrt{3}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{3}}{2*1}=\frac{10+2\sqrt{3}}{2}
$