2x^2+6x-1927=0

Solution for 2x^2+6x-1927=0 equation:

2x^2+6x-1927=0

a = 2; b = 6; c = -1927;
Δ = b2-4ac
Δ = 62-4·2·(-1927)
Δ = 15452
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:

$\sqrt{\Delta}=\sqrt{15452}=\sqrt{4*3863}=\sqrt{4}*\sqrt{3863}=2\sqrt{3863}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{3863}}{2*2}=\frac{-6-2\sqrt{3863}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{3863}}{2*2}=\frac{-6+2\sqrt{3863}}{4}
$