(2x-1)*(3x+1)=21

Solution for (2x-1)*(3x+1)=21 equation:

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

We move all terms to the left:

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

We multiply parentheses ..

(+6x^2+2x-3x-1)-21=0

We get rid of parentheses

6x^2+2x-3x-1-21=0

We add all the numbers together, and all the variables

6x^2-1x-22=0

a = 6; b = -1; c = -22;
Δ = b2-4ac
Δ = -12-4·6·(-22)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-23}{2*6}=\frac{-22}{12}
=-1+5/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+23}{2*6}=\frac{24}{12}
=2
$