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

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

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

We move all terms to the left:

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

We multiply parentheses

6x-(5(2x-4)4x)+10=0


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

5(2x-4)4x

We multiply parentheses

40x^2-80x

Back to the equation:

-(40x^2-80x)


We get rid of parentheses

-40x^2+6x+80x+10=0

We add all the numbers together, and all the variables

-40x^2+86x+10=0

a = -40; b = 86; c = +10;
Δ = b2-4ac
Δ = 862-4·(-40)·10
Δ = 8996
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8996}=\sqrt{4*2249}=\sqrt{4}*\sqrt{2249}=2\sqrt{2249}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(86)-2\sqrt{2249}}{2*-40}=\frac{-86-2\sqrt{2249}}{-80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(86)+2\sqrt{2249}}{2*-40}=\frac{-86+2\sqrt{2249}}{-80}
$