(2x-3)(3x-4)+(2x-3)(x+1)=0

Solution for (2x-3)(3x-4)+(2x-3)(x+1)=0 equation:

(2x-3)(3x-4)+(2x-3)(x+1)=0

We multiply parentheses ..

(+6x^2-8x-9x+12)+(2x-3)(x+1)=0

We get rid of parentheses

6x^2-8x-9x+(2x-3)(x+1)+12=0

We multiply parentheses ..

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

We add all the numbers together, and all the variables

6x^2+(+2x^2+2x-3x-3)-17x+12=0

We get rid of parentheses

6x^2+2x^2+2x-3x-17x-3+12=0

We add all the numbers together, and all the variables

8x^2-18x+9=0

a = 8; b = -18; c = +9;
Δ = b2-4ac
Δ = -182-4·8·9
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6}{2*8}=\frac{12}{16}
=3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6}{2*8}=\frac{24}{16}
=1+1/2
$