2x(12-1/2)=2(4-2x)+20

Solution for 2x(12-1/2)=2(4-2x)+20 equation:

2x(12-1/2)=2(4-2x)+20

We move all terms to the left:

2x(12-1/2)-(2(4-2x)+20)=0

We add all the numbers together, and all the variables

2x(-1/2+12)-(2(-2x+4)+20)=0

We multiply parentheses

-2x^2+24x-(2(-2x+4)+20)=0


We calculate terms in parentheses: -(2(-2x+4)+20), so:

2(-2x+4)+20

We multiply parentheses

-4x+8+20

We add all the numbers together, and all the variables

-4x+28

Back to the equation:

-(-4x+28)


We get rid of parentheses

-2x^2+24x+4x-28=0

We add all the numbers together, and all the variables

-2x^2+28x-28=0

a = -2; b = 28; c = -28;
Δ = b2-4ac
Δ = 282-4·(-2)·(-28)
Δ = 560
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:

$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{35}}{2*-2}=\frac{-28-4\sqrt{35}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{35}}{2*-2}=\frac{-28+4\sqrt{35}}{-4}
$