66=(3x+1)(x-4)

Solution for 66=(3x+1)(x-4) equation:

66=(3x+1)(x-4)

We move all terms to the left:

66-((3x+1)(x-4))=0

We multiply parentheses ..

-((+3x^2-12x+x-4))+66=0


We calculate terms in parentheses: -((+3x^2-12x+x-4)), so:

(+3x^2-12x+x-4)

We get rid of parentheses

3x^2-12x+x-4

We add all the numbers together, and all the variables

3x^2-11x-4

Back to the equation:

-(3x^2-11x-4)


We get rid of parentheses

-3x^2+11x+4+66=0

We add all the numbers together, and all the variables

-3x^2+11x+70=0

a = -3; b = 11; c = +70;
Δ = b2-4ac
Δ = 112-4·(-3)·70
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-31}{2*-3}=\frac{-42}{-6}
=+7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+31}{2*-3}=\frac{20}{-6}
=-3+1/3
$