20(3x/4)=20(4x/5)+(.25)20

Solution for 20(3x/4)=20(4x/5)+(.25)20 equation:

20(3x/4)=20(4x/5)+(.25)20

We move all terms to the left:

20(3x/4)-(20(4x/5)+(.25)20)=0

We add all the numbers together, and all the variables

20(+3x/4)-(20(+4x/5)+(0.25)20)=0

We multiply parentheses

60x-(20(+4x/5)+(0.25)20)=0

We multiply all the terms by the denominator


60x*5)+(0.25)20)-(20(+4x=0

We add all the numbers together, and all the variables

4x+60x*5)+(0.25)20)-(20(=0

Wy multiply elements

300x^2+4x=0

a = 300; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·300·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*300}=\frac{-8}{600}
=-1/75
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*300}=\frac{0}{600}
=0
$