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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+3x^2-12x+x-4)-1=0

We get rid of parentheses

3x^2-12x+x-4-1=0

We add all the numbers together, and all the variables

3x^2-11x-5=0

a = 3; b = -11; c = -5;
Δ = b2-4ac
Δ = -112-4·3·(-5)
Δ = 181
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{181}}{2*3}=\frac{11-\sqrt{181}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{181}}{2*3}=\frac{11+\sqrt{181}}{6}
$