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

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

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

We move all terms to the left:

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

We multiply parentheses

20x-6x-((2x-5)(5x-1)-3)-4+15=0

We multiply parentheses ..

-((+10x^2-2x-25x+5)-3)+20x-6x-4+15=0


We calculate terms in parentheses: -((+10x^2-2x-25x+5)-3), so:

(+10x^2-2x-25x+5)-3

We get rid of parentheses

10x^2-2x-25x+5-3

We add all the numbers together, and all the variables

10x^2-27x+2

Back to the equation:

-(10x^2-27x+2)


We add all the numbers together, and all the variables

14x-(10x^2-27x+2)+11=0

We get rid of parentheses

-10x^2+14x+27x-2+11=0

We add all the numbers together, and all the variables

-10x^2+41x+9=0

a = -10; b = 41; c = +9;
Δ = b2-4ac
Δ = 412-4·(-10)·9
Δ = 2041
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-\sqrt{2041}}{2*-10}=\frac{-41-\sqrt{2041}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+\sqrt{2041}}{2*-10}=\frac{-41+\sqrt{2041}}{-20}
$