r^2+28r+56=0

Solution for r^2+28r+56=0 equation:

r^2+28r+56=0

a = 1; b = 28; c = +56;
Δ = b2-4ac
Δ = 282-4·1·56
Δ = 560
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:

$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{35}}{2*1}=\frac{-28-4\sqrt{35}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{35}}{2*1}=\frac{-28+4\sqrt{35}}{2}
$