2x^2+4x-94=0

Solution for 2x^2+4x-94=0 equation:

2x^2+4x-94=0

a = 2; b = 4; c = -94;
Δ = b2-4ac
Δ = 42-4·2·(-94)
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16\sqrt{3}}{2*2}=\frac{-4-16\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16\sqrt{3}}{2*2}=\frac{-4+16\sqrt{3}}{4}
$