2x^2+x-208=0

Solution for 2x^2+x-208=0 equation:

2x^2+x-208=0

a = 2; b = 1; c = -208;
Δ = b2-4ac
Δ = 12-4·2·(-208)
Δ = 1665
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:

$\sqrt{\Delta}=\sqrt{1665}=\sqrt{9*185}=\sqrt{9}*\sqrt{185}=3\sqrt{185}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{185}}{2*2}=\frac{-1-3\sqrt{185}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{185}}{2*2}=\frac{-1+3\sqrt{185}}{4}
$