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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+12x^2-20x+9x-15)-1x=0

We get rid of parentheses

12x^2-20x+9x-1x-15=0

We add all the numbers together, and all the variables

12x^2-12x-15=0

a = 12; b = -12; c = -15;
Δ = b2-4ac
Δ = -122-4·12·(-15)
Δ = 864
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:

$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12\sqrt{6}}{2*12}=\frac{12-12\sqrt{6}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12\sqrt{6}}{2*12}=\frac{12+12\sqrt{6}}{24}
$