2(3x+7)=4(2x+2)32X=

Solution for 2(3x+7)=4(2x+2)32X= equation:

2(3x+7)=4(2x+2)32x=

We move all terms to the left:

2(3x+7)-(4(2x+2)32x)=0

We multiply parentheses

6x-(4(2x+2)32x)+14=0


We calculate terms in parentheses: -(4(2x+2)32x), so:

4(2x+2)32x

We multiply parentheses

256x^2+256x

Back to the equation:

-(256x^2+256x)


We get rid of parentheses

-256x^2+6x-256x+14=0

We add all the numbers together, and all the variables

-256x^2-250x+14=0

a = -256; b = -250; c = +14;
Δ = b2-4ac
Δ = -2502-4·(-256)·14
Δ = 76836
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:

$\sqrt{\Delta}=\sqrt{76836}=\sqrt{4*19209}=\sqrt{4}*\sqrt{19209}=2\sqrt{19209}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-250)-2\sqrt{19209}}{2*-256}=\frac{250-2\sqrt{19209}}{-512}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-250)+2\sqrt{19209}}{2*-256}=\frac{250+2\sqrt{19209}}{-512}
$