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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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

We get rid of parentheses

3x^2-3x+x-1-64=0

We add all the numbers together, and all the variables

3x^2-2x-65=0

a = 3; b = -2; c = -65;
Δ = b2-4ac
Δ = -22-4·3·(-65)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-28}{2*3}=\frac{-26}{6}
=-4+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+28}{2*3}=\frac{30}{6}
=5
$