(4x-3)(3x+2)=5x+1

Solution for (4x-3)(3x+2)=5x+1 equation:

(4x-3)(3x+2)=5x+1

We move all terms to the left:

(4x-3)(3x+2)-(5x+1)=0

We get rid of parentheses

(4x-3)(3x+2)-5x-1=0

We multiply parentheses ..

(+12x^2+8x-9x-6)-5x-1=0

We get rid of parentheses

12x^2+8x-9x-5x-6-1=0

We add all the numbers together, and all the variables

12x^2-6x-7=0

a = 12; b = -6; c = -7;
Δ = b2-4ac
Δ = -62-4·12·(-7)
Δ = 372
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:

$\sqrt{\Delta}=\sqrt{372}=\sqrt{4*93}=\sqrt{4}*\sqrt{93}=2\sqrt{93}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{93}}{2*12}=\frac{6-2\sqrt{93}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{93}}{2*12}=\frac{6+2\sqrt{93}}{24}
$