(3x+1)(x+1)=21

Solution for (3x+1)(x+1)=21 equation:

(3x+1)(x+1)=21

We move all terms to the left:

(3x+1)(x+1)-(21)=0

We multiply parentheses ..

(+3x^2+3x+x+1)-21=0

We get rid of parentheses

3x^2+3x+x+1-21=0

We add all the numbers together, and all the variables

3x^2+4x-20=0

a = 3; b = 4; c = -20;
Δ = b2-4ac
Δ = 42-4·3·(-20)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*3}=\frac{-20}{6}
=-3+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*3}=\frac{12}{6}
=2
$