(4x-5)(3x+2)-3x=5

Solution for (4x-5)(3x+2)-3x=5 equation:

(4x-5)(3x+2)-3x=5

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+12x^2+8x-15x-10)-3x-5=0

We get rid of parentheses

12x^2+8x-15x-3x-10-5=0

We add all the numbers together, and all the variables

12x^2-10x-15=0

a = 12; b = -10; c = -15;
Δ = b2-4ac
Δ = -102-4·12·(-15)
Δ = 820
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{820}=\sqrt{4*205}=\sqrt{4}*\sqrt{205}=2\sqrt{205}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{205}}{2*12}=\frac{10-2\sqrt{205}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{205}}{2*12}=\frac{10+2\sqrt{205}}{24}
$