3/3x+100-20=7x+800

Solution for 3/3x+100-20=7x+800 equation:

3/3x+100-20=7x+800

We move all terms to the left:

3/3x+100-20-(7x+800)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

3/3x-(7x+800)+80=0

We get rid of parentheses

3/3x-7x-800+80=0

We multiply all the terms by the denominator


-7x*3x-800*3x+80*3x+3=0

Wy multiply elements

-21x^2-2400x+240x+3=0

We add all the numbers together, and all the variables

-21x^2-2160x+3=0

a = -21; b = -2160; c = +3;
Δ = b2-4ac
Δ = -21602-4·(-21)·3
Δ = 4665852
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:

$\sqrt{\Delta}=\sqrt{4665852}=\sqrt{36*129607}=\sqrt{36}*\sqrt{129607}=6\sqrt{129607}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2160)-6\sqrt{129607}}{2*-21}=\frac{2160-6\sqrt{129607}}{-42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2160)+6\sqrt{129607}}{2*-21}=\frac{2160+6\sqrt{129607}}{-42}
$