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

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

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

We move all terms to the left:

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

We multiply parentheses

12x^2-80x-(5x(2x+4))=0


We calculate terms in parentheses: -(5x(2x+4)), so:

5x(2x+4)

We multiply parentheses

10x^2+20x

Back to the equation:

-(10x^2+20x)


We get rid of parentheses

12x^2-10x^2-80x-20x=0

We add all the numbers together, and all the variables

2x^2-100x=0

a = 2; b = -100; c = 0;
Δ = b2-4ac
Δ = -1002-4·2·0
Δ = 10000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10000}=100$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-100}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+100}{2*2}=\frac{200}{4}
=50
$