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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-6x^2+20x+8x-3=0

We add all the numbers together, and all the variables

-6x^2+28x-3=0

a = -6; b = 28; c = -3;
Δ = b2-4ac
Δ = 282-4·(-6)·(-3)
Δ = 712
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:

$\sqrt{\Delta}=\sqrt{712}=\sqrt{4*178}=\sqrt{4}*\sqrt{178}=2\sqrt{178}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{178}}{2*-6}=\frac{-28-2\sqrt{178}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{178}}{2*-6}=\frac{-28+2\sqrt{178}}{-12}
$