(3x-1)(x-4)=16

Solution for (3x-1)(x-4)=16 equation:

(3x-1)(x-4)=16

We move all terms to the left:

(3x-1)(x-4)-(16)=0

We multiply parentheses ..

(+3x^2-12x-1x+4)-16=0

We get rid of parentheses

3x^2-12x-1x+4-16=0

We add all the numbers together, and all the variables

3x^2-13x-12=0

a = 3; b = -13; c = -12;
Δ = b2-4ac
Δ = -132-4·3·(-12)
Δ = 313
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{-(-13)-\sqrt{313}}{2*3}=\frac{13-\sqrt{313}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{313}}{2*3}=\frac{13+\sqrt{313}}{6}
$