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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+2x^2+3x-8x-12)-((4x-3)(3x+4)+2)=0


We calculate terms in parentheses: -((4x-3)(3x+4)+2), so:

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

We multiply parentheses ..

(+12x^2+16x-9x-12)+2

We get rid of parentheses

12x^2+16x-9x-12+2

We add all the numbers together, and all the variables

12x^2+7x-10

Back to the equation:

-(12x^2+7x-10)


We get rid of parentheses

2x^2-12x^2+3x-8x-7x-12+10=0

We add all the numbers together, and all the variables

-10x^2-12x-2=0

a = -10; b = -12; c = -2;
Δ = b2-4ac
Δ = -122-4·(-10)·(-2)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8}{2*-10}=\frac{4}{-20}
=-1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8}{2*-10}=\frac{20}{-20}
=-1
$