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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4x^2+8x-1x-2)-(-1x+3)(4x-1)=0

We get rid of parentheses

4x^2+8x-1x-(-1x+3)(4x-1)-2=0

We multiply parentheses ..

4x^2-(-4x^2+x+12x-3)+8x-1x-2=0

We add all the numbers together, and all the variables

4x^2-(-4x^2+x+12x-3)+7x-2=0

We get rid of parentheses

4x^2+4x^2-x-12x+7x+3-2=0

We add all the numbers together, and all the variables

8x^2-6x+1=0

a = 8; b = -6; c = +1;
Δ = b2-4ac
Δ = -62-4·8·1
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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