2x^2+4x-16=4

Solution for 2x^2+4x-16=4 equation:

2x^2+4x-16=4

We move all terms to the left:

2x^2+4x-16-(4)=0

We add all the numbers together, and all the variables

2x^2+4x-20=0

a = 2; b = 4; c = -20;
Δ = b2-4ac
Δ = 42-4·2·(-20)
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{11}}{2*2}=\frac{-4-4\sqrt{11}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{11}}{2*2}=\frac{-4+4\sqrt{11}}{4}
$