(5x)(4x+8)=(4x)(6x-10)

Solution for (5x)(4x+8)=(4x)(6x-10) equation:

(5x)(4x+8)=(4x)(6x-10)

We move all terms to the left:

(5x)(4x+8)-((4x)(6x-10))=0

We multiply parentheses

20x^2+40x-(4x(6x-10))=0


We calculate terms in parentheses: -(4x(6x-10)), so:

4x(6x-10)

We multiply parentheses

24x^2-40x

Back to the equation:

-(24x^2-40x)


We get rid of parentheses

20x^2-24x^2+40x+40x=0

We add all the numbers together, and all the variables

-4x^2+80x=0

a = -4; b = 80; c = 0;
Δ = b2-4ac
Δ = 802-4·(-4)·0
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-80}{2*-4}=\frac{-160}{-8}
=+20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+80}{2*-4}=\frac{0}{-8}
=0
$