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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4x^2+20x+3x-15)-((x-5)(3x+1))=0


We calculate terms in parentheses: -((x-5)(3x+1)), so:

(x-5)(3x+1)

We multiply parentheses ..

(+3x^2+x-15x-5)

We get rid of parentheses

3x^2+x-15x-5

We add all the numbers together, and all the variables

3x^2-14x-5

Back to the equation:

-(3x^2-14x-5)


We get rid of parentheses

-4x^2-3x^2+20x+3x+14x-15+5=0

We add all the numbers together, and all the variables

-7x^2+37x-10=0

a = -7; b = 37; c = -10;
Δ = b2-4ac
Δ = 372-4·(-7)·(-10)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(37)-33}{2*-7}=\frac{-70}{-14}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(37)+33}{2*-7}=\frac{-4}{-14}
=2/7
$