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

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

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

We multiply parentheses ..

4x^2+(+6x^2+4x-3x-2)-4x+1=0

We get rid of parentheses

4x^2+6x^2+4x-3x-4x-2+1=0

We add all the numbers together, and all the variables

10x^2-3x-1=0

a = 10; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·10·(-1)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*10}=\frac{-4}{20}
=-1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*10}=\frac{10}{20}
=1/2
$