r^2-12r-28=-10

Solution for r^2-12r-28=-10 equation:

r^2-12r-28=-10

We move all terms to the left:

r^2-12r-28-(-10)=0

We add all the numbers together, and all the variables

r^2-12r-18=0

a = 1; b = -12; c = -18;
Δ = b2-4ac
Δ = -122-4·1·(-18)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{6}}{2*1}=\frac{12-6\sqrt{6}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{6}}{2*1}=\frac{12+6\sqrt{6}}{2}
$