(3x-2)(x+4)=1

Solution for (3x-2)(x+4)=1 equation:

(3x-2)(x+4)=1

We move all terms to the left:

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

We multiply parentheses ..

(+3x^2+12x-2x-8)-1=0

We get rid of parentheses

3x^2+12x-2x-8-1=0

We add all the numbers together, and all the variables

3x^2+10x-9=0

a = 3; b = 10; c = -9;
Δ = b2-4ac
Δ = 102-4·3·(-9)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{13}}{2*3}=\frac{-10-4\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{13}}{2*3}=\frac{-10+4\sqrt{13}}{6}
$