66=4(2x-3)+2(x+4)x=

Solution for 66=4(2x-3)+2(x+4)x= equation:

66=4(2x-3)+2(x+4)x=

We move all terms to the left:

66-(4(2x-3)+2(x+4)x)=0


We calculate terms in parentheses: -(4(2x-3)+2(x+4)x), so:

4(2x-3)+2(x+4)x

We multiply parentheses

2x^2+8x+8x-12

We add all the numbers together, and all the variables

2x^2+16x-12

Back to the equation:

-(2x^2+16x-12)


We get rid of parentheses

-2x^2-16x+12+66=0

We add all the numbers together, and all the variables

-2x^2-16x+78=0

a = -2; b = -16; c = +78;
Δ = b2-4ac
Δ = -162-4·(-2)·78
Δ = 880
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:

$\sqrt{\Delta}=\sqrt{880}=\sqrt{16*55}=\sqrt{16}*\sqrt{55}=4\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{55}}{2*-2}=\frac{16-4\sqrt{55}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{55}}{2*-2}=\frac{16+4\sqrt{55}}{-4}
$