x^2+56x+140=0

Solution for x^2+56x+140=0 equation:

x^2+56x+140=0

a = 1; b = 56; c = +140;
Δ = b2-4ac
Δ = 562-4·1·140
Δ = 2576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2576}=\sqrt{16*161}=\sqrt{16}*\sqrt{161}=4\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-4\sqrt{161}}{2*1}=\frac{-56-4\sqrt{161}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+4\sqrt{161}}{2*1}=\frac{-56+4\sqrt{161}}{2}
$