X^2+560x-5400=0

Solution for X^2+560x-5400=0 equation:

X^2+560X-5400=0

a = 1; b = 560; c = -5400;
Δ = b2-4ac
Δ = 5602-4·1·(-5400)
Δ = 335200
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:

$\sqrt{\Delta}=\sqrt{335200}=\sqrt{400*838}=\sqrt{400}*\sqrt{838}=20\sqrt{838}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(560)-20\sqrt{838}}{2*1}=\frac{-560-20\sqrt{838}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(560)+20\sqrt{838}}{2*1}=\frac{-560+20\sqrt{838}}{2}
$