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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -(4x(x-1)), so:

4x(x-1)

We multiply parentheses

4x^2-4x

Back to the equation:

-(4x^2-4x)


We add all the numbers together, and all the variables

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

We get rid of parentheses

-4x^2+8x+4x-1=0

We add all the numbers together, and all the variables

-4x^2+12x-1=0

a = -4; b = 12; c = -1;
Δ = b2-4ac
Δ = 122-4·(-4)·(-1)
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{2}}{2*-4}=\frac{-12-8\sqrt{2}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{2}}{2*-4}=\frac{-12+8\sqrt{2}}{-8}
$