(5x-4)(x+4)=(4x-5)(x+8)

Solution for (5x-4)(x+4)=(4x-5)(x+8) equation:

(5x-4)(x+4)=(4x-5)(x+8)

We move all terms to the left:

(5x-4)(x+4)-((4x-5)(x+8))=0

We multiply parentheses ..

(+5x^2+20x-4x-16)-((4x-5)(x+8))=0


We calculate terms in parentheses: -((4x-5)(x+8)), so:

(4x-5)(x+8)

We multiply parentheses ..

(+4x^2+32x-5x-40)

We get rid of parentheses

4x^2+32x-5x-40

We add all the numbers together, and all the variables

4x^2+27x-40

Back to the equation:

-(4x^2+27x-40)


We get rid of parentheses

5x^2-4x^2+20x-4x-27x-16+40=0

We add all the numbers together, and all the variables

x^2-11x+24=0

a = 1; b = -11; c = +24;
Δ = b2-4ac
Δ = -112-4·1·24
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*1}=\frac{6}{2}
=3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*1}=\frac{16}{2}
=8
$