2x^2+4x-78=108

Solution for 2x^2+4x-78=108 equation:

2x^2+4x-78=108

We move all terms to the left:

2x^2+4x-78-(108)=0

We add all the numbers together, and all the variables

2x^2+4x-186=0

a = 2; b = 4; c = -186;
Δ = b2-4ac
Δ = 42-4·2·(-186)
Δ = 1504
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:

$\sqrt{\Delta}=\sqrt{1504}=\sqrt{16*94}=\sqrt{16}*\sqrt{94}=4\sqrt{94}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{94}}{2*2}=\frac{-4-4\sqrt{94}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{94}}{2*2}=\frac{-4+4\sqrt{94}}{4}
$