5x^2+40x+56=0

Solution for 5x^2+40x+56=0 equation:

5x^2+40x+56=0

a = 5; b = 40; c = +56;
Δ = b2-4ac
Δ = 402-4·5·56
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{30}}{2*5}=\frac{-40-4\sqrt{30}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{30}}{2*5}=\frac{-40+4\sqrt{30}}{10}
$