2x^2+x-6400=0

Solution for 2x^2+x-6400=0 equation:

2x^2+x-6400=0

a = 2; b = 1; c = -6400;
Δ = b2-4ac
Δ = 12-4·2·(-6400)
Δ = 51201
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:

$\sqrt{\Delta}=\sqrt{51201}=\sqrt{9*5689}=\sqrt{9}*\sqrt{5689}=3\sqrt{5689}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{5689}}{2*2}=\frac{-1-3\sqrt{5689}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{5689}}{2*2}=\frac{-1+3\sqrt{5689}}{4}
$