2x^2-4x-99=0

Solution for 2x^2-4x-99=0 equation:

2x^2-4x-99=0

a = 2; b = -4; c = -99;
Δ = b2-4ac
Δ = -42-4·2·(-99)
Δ = 808
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:

$\sqrt{\Delta}=\sqrt{808}=\sqrt{4*202}=\sqrt{4}*\sqrt{202}=2\sqrt{202}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{202}}{2*2}=\frac{4-2\sqrt{202}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{202}}{2*2}=\frac{4+2\sqrt{202}}{4}
$