(2x-4)+(3x^2+2x-4)=0

Solution for (2x-4)+(3x^2+2x-4)=0 equation:

(2x-4)+(3x^2+2x-4)=0

We get rid of parentheses

3x^2+2x+2x-4-4=0

We add all the numbers together, and all the variables

3x^2+4x-8=0

a = 3; b = 4; c = -8;
Δ = b2-4ac
Δ = 42-4·3·(-8)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{7}}{2*3}=\frac{-4-4\sqrt{7}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{7}}{2*3}=\frac{-4+4\sqrt{7}}{6}
$