3x(x-1)=7x^2-4x

Solution for 3x(x-1)=7x^2-4x equation:

3x(x-1)=7x^2-4x

We move all terms to the left:

3x(x-1)-(7x^2-4x)=0

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4x^2+x=0

a = -4; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-4)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-4}=\frac{-2}{-8}
=1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-4}=\frac{0}{-8}
=0
$