4x(3x-20)=5x(2x+8)

Solution for 4x(3x-20)=5x(2x+8) equation:

4x(3x-20)=5x(2x+8)

We move all terms to the left:

4x(3x-20)-(5x(2x+8))=0

We multiply parentheses

12x^2-80x-(5x(2x+8))=0


We calculate terms in parentheses: -(5x(2x+8)), so:

5x(2x+8)

We multiply parentheses

10x^2+40x

Back to the equation:

-(10x^2+40x)


We get rid of parentheses

12x^2-10x^2-80x-40x=0

We add all the numbers together, and all the variables

2x^2-120x=0

a = 2; b = -120; c = 0;
Δ = b2-4ac
Δ = -1202-4·2·0
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*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+120}{2*2}=\frac{240}{4}
=60
$