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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+6x^2+4x+9x+6)-5=0

We get rid of parentheses

6x^2+4x+9x+6-5=0

We add all the numbers together, and all the variables

6x^2+13x+1=0

a = 6; b = 13; c = +1;
Δ = b2-4ac
Δ = 132-4·6·1
Δ = 145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{145}}{2*6}=\frac{-13-\sqrt{145}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{145}}{2*6}=\frac{-13+\sqrt{145}}{12}
$