(3x+1)(2x-2)=64x=

Solution for (3x+1)(2x-2)=64x= equation:

(3x+1)(2x-2)=64x=

We move all terms to the left:

(3x+1)(2x-2)-(64x)=0

We add all the numbers together, and all the variables

-64x+(3x+1)(2x-2)=0

We multiply parentheses ..

(+6x^2-6x+2x-2)-64x=0

We get rid of parentheses

6x^2-6x+2x-64x-2=0

We add all the numbers together, and all the variables

6x^2-68x-2=0

a = 6; b = -68; c = -2;
Δ = b2-4ac
Δ = -682-4·6·(-2)
Δ = 4672
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:

$\sqrt{\Delta}=\sqrt{4672}=\sqrt{64*73}=\sqrt{64}*\sqrt{73}=8\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-68)-8\sqrt{73}}{2*6}=\frac{68-8\sqrt{73}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-68)+8\sqrt{73}}{2*6}=\frac{68+8\sqrt{73}}{12}
$