3x(x-4)+x=20

Solution for 3x(x-4)+x=20 equation:

3x(x-4)+x=20

We move all terms to the left:

3x(x-4)+x-(20)=0

We add all the numbers together, and all the variables

x+3x(x-4)-20=0

We multiply parentheses

3x^2+x-12x-20=0

We add all the numbers together, and all the variables

3x^2-11x-20=0

a = 3; b = -11; c = -20;
Δ = b2-4ac
Δ = -112-4·3·(-20)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-19}{2*3}=\frac{-8}{6}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+19}{2*3}=\frac{30}{6}
=5
$