(2x^2-2x+4)-2(3x+5)=0

Solution for (2x^2-2x+4)-2(3x+5)=0 equation:

(2x^2-2x+4)-2(3x+5)=0

We multiply parentheses

(2x^2-2x+4)-6x-10=0

We get rid of parentheses

2x^2-2x-6x+4-10=0

We add all the numbers together, and all the variables

2x^2-8x-6=0

a = 2; b = -8; c = -6;
Δ = b2-4ac
Δ = -82-4·2·(-6)
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{-(-8)-4\sqrt{7}}{2*2}=\frac{8-4\sqrt{7}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{7}}{2*2}=\frac{8+4\sqrt{7}}{4}
$