2x^2+4x-541=0

Solution for 2x^2+4x-541=0 equation:

2x^2+4x-541=0

a = 2; b = 4; c = -541;
Δ = b2-4ac
Δ = 42-4·2·(-541)
Δ = 4344
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:

$\sqrt{\Delta}=\sqrt{4344}=\sqrt{4*1086}=\sqrt{4}*\sqrt{1086}=2\sqrt{1086}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{1086}}{2*2}=\frac{-4-2\sqrt{1086}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{1086}}{2*2}=\frac{-4+2\sqrt{1086}}{4}
$