4x(6x-20)=5x(4x+8)

Solution for 4x(6x-20)=5x(4x+8) equation:

4x(6x-20)=5x(4x+8)

We move all terms to the left:

4x(6x-20)-(5x(4x+8))=0

We multiply parentheses

24x^2-80x-(5x(4x+8))=0


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

5x(4x+8)

We multiply parentheses

20x^2+40x

Back to the equation:

-(20x^2+40x)


We get rid of parentheses

24x^2-20x^2-80x-40x=0

We add all the numbers together, and all the variables

4x^2-120x=0

a = 4; b = -120; c = 0;
Δ = b2-4ac
Δ = -1202-4·4·0
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-120}{2*4}=\frac{0}{8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+120}{2*4}=\frac{240}{8}
=30
$