(4-5x)(4x-5)=(10x-3)(7+2x)

Solution for (4-5x)(4x-5)=(10x-3)(7+2x) equation:

(4-5x)(4x-5)=(10x-3)(7+2x)

We move all terms to the left:

(4-5x)(4x-5)-((10x-3)(7+2x))=0

We add all the numbers together, and all the variables

(-5x+4)(4x-5)-((10x-3)(2x+7))=0

We multiply parentheses ..

(-20x^2+25x+16x-20)-((10x-3)(2x+7))=0


We calculate terms in parentheses: -((10x-3)(2x+7)), so:

(10x-3)(2x+7)

We multiply parentheses ..

(+20x^2+70x-6x-21)

We get rid of parentheses

20x^2+70x-6x-21

We add all the numbers together, and all the variables

20x^2+64x-21

Back to the equation:

-(20x^2+64x-21)


We get rid of parentheses

-20x^2-20x^2+25x+16x-64x-20+21=0

We add all the numbers together, and all the variables

-40x^2-23x+1=0

a = -40; b = -23; c = +1;
Δ = b2-4ac
Δ = -232-4·(-40)·1
Δ = 689
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{-(-23)-\sqrt{689}}{2*-40}=\frac{23-\sqrt{689}}{-80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{689}}{2*-40}=\frac{23+\sqrt{689}}{-80}
$