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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((x-1)(-3x+4)), so:

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

We multiply parentheses ..

(-3x^2+4x+3x-4)

We get rid of parentheses

-3x^2+4x+3x-4

We add all the numbers together, and all the variables

-3x^2+7x-4

Back to the equation:

-(-3x^2+7x-4)


We get rid of parentheses

2x^2+3x^2+5x-2x-7x-5+4=0

We add all the numbers together, and all the variables

5x^2-4x-1=0

a = 5; b = -4; c = -1;
Δ = b2-4ac
Δ = -42-4·5·(-1)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6}{2*5}=\frac{-2}{10}
=-1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*5}=\frac{10}{10}
=1
$