5x^2+40x-560=0

Solution for 5x^2+40x-560=0 equation:

5x^2+40x-560=0

a = 5; b = 40; c = -560;
Δ = b2-4ac
Δ = 402-4·5·(-560)
Δ = 12800
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:

$\sqrt{\Delta}=\sqrt{12800}=\sqrt{6400*2}=\sqrt{6400}*\sqrt{2}=80\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-80\sqrt{2}}{2*5}=\frac{-40-80\sqrt{2}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+80\sqrt{2}}{2*5}=\frac{-40+80\sqrt{2}}{10}
$