(2x+1)(3x-2)=5

Solution for (2x+1)(3x-2)=5 equation:

(2x+1)(3x-2)=5

We move all terms to the left:

(2x+1)(3x-2)-(5)=0

We multiply parentheses ..

(+6x^2-4x+3x-2)-5=0

We get rid of parentheses

6x^2-4x+3x-2-5=0

We add all the numbers together, and all the variables

6x^2-1x-7=0

a = 6; b = -1; c = -7;
Δ = b2-4ac
Δ = -12-4·6·(-7)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-13}{2*6}=\frac{-12}{12}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+13}{2*6}=\frac{14}{12}
=1+1/6
$