1/3x+45=3x-35

Solution for 1/3x+45=3x-35 equation:

1/3x+45=3x-35

We move all terms to the left:

1/3x+45-(3x-35)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We get rid of parentheses

1/3x-3x+35+45=0

We multiply all the terms by the denominator


-3x*3x+35*3x+45*3x+1=0

Wy multiply elements

-9x^2+105x+135x+1=0

We add all the numbers together, and all the variables

-9x^2+240x+1=0

a = -9; b = 240; c = +1;
Δ = b2-4ac
Δ = 2402-4·(-9)·1
Δ = 57636
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:

$\sqrt{\Delta}=\sqrt{57636}=\sqrt{36*1601}=\sqrt{36}*\sqrt{1601}=6\sqrt{1601}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(240)-6\sqrt{1601}}{2*-9}=\frac{-240-6\sqrt{1601}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(240)+6\sqrt{1601}}{2*-9}=\frac{-240+6\sqrt{1601}}{-18}
$