r^2+12r+28=0

Solution for r^2+12r+28=0 equation:

r^2+12r+28=0

a = 1; b = 12; c = +28;
Δ = b2-4ac
Δ = 122-4·1·28
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{2}}{2*1}=\frac{-12-4\sqrt{2}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{2}}{2*1}=\frac{-12+4\sqrt{2}}{2}
$