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

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

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(2x(3x-4)), so:

2x(3x-4)

We multiply parentheses

6x^2-8x

Back to the equation:

-(6x^2-8x)


We get rid of parentheses

8x^2-6x^2+20x+8x=0

We add all the numbers together, and all the variables

2x^2+28x=0

a = 2; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·2·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*2}=\frac{-56}{4}
=-14
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*2}=\frac{0}{4}
=0
$