(3x+1)(x-2)=20

Solution for (3x+1)(x-2)=20 equation:

(3x+1)(x-2)=20

We move all terms to the left:

(3x+1)(x-2)-(20)=0

We multiply parentheses ..

(+3x^2-6x+x-2)-20=0

We get rid of parentheses

3x^2-6x+x-2-20=0

We add all the numbers together, and all the variables

3x^2-5x-22=0

a = 3; b = -5; c = -22;
Δ = b2-4ac
Δ = -52-4·3·(-22)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-17}{2*3}=\frac{-12}{6}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+17}{2*3}=\frac{22}{6}
=3+2/3
$