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

Solution for (2x+1)(5x-4)-(3x-2)2=0 equation:

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

We multiply parentheses

(2x+1)(5x-4)-6x+4=0

We multiply parentheses ..

(+10x^2-8x+5x-4)-6x+4=0

We get rid of parentheses

10x^2-8x+5x-6x-4+4=0

We add all the numbers together, and all the variables

10x^2-9x=0

a = 10; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·10·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*10}=\frac{0}{20}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*10}=\frac{18}{20}
=9/10
$