(2x+3)(3x-4)=3

Solution for (2x+3)(3x-4)=3 equation:

(2x+3)(3x-4)=3

We move all terms to the left:

(2x+3)(3x-4)-(3)=0

We multiply parentheses ..

(+6x^2-8x+9x-12)-3=0

We get rid of parentheses

6x^2-8x+9x-12-3=0

We add all the numbers together, and all the variables

6x^2+x-15=0

a = 6; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·6·(-15)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*6}=\frac{-20}{12}
=-1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*6}=\frac{18}{12}
=1+1/2
$