1/3x+(180-x)=98

Solution for 1/3x+(180-x)=98 equation:

1/3x+(180-x)=98

We move all terms to the left:

1/3x+(180-x)-(98)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

1/3x+(-1x+180)-98=0

We get rid of parentheses

1/3x-1x+180-98=0

We multiply all the terms by the denominator


-1x*3x+180*3x-98*3x+1=0

Wy multiply elements

-3x^2+540x-294x+1=0

We add all the numbers together, and all the variables

-3x^2+246x+1=0

a = -3; b = 246; c = +1;
Δ = b2-4ac
Δ = 2462-4·(-3)·1
Δ = 60528
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:

$\sqrt{\Delta}=\sqrt{60528}=\sqrt{16*3783}=\sqrt{16}*\sqrt{3783}=4\sqrt{3783}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(246)-4\sqrt{3783}}{2*-3}=\frac{-246-4\sqrt{3783}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(246)+4\sqrt{3783}}{2*-3}=\frac{-246+4\sqrt{3783}}{-6}
$