x-20+x-10+1/3x+40=360

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Solution for x-20+x-10+1/3x+40=360 equation:



x-20+x-10+1/3x+40=360
We move all terms to the left:
x-20+x-10+1/3x+40-(360)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
We add all the numbers together, and all the variables
2x+1/3x-350=0
We multiply all the terms by the denominator
2x*3x-350*3x+1=0
Wy multiply elements
6x^2-1050x+1=0
a = 6; b = -1050; c = +1;
Δ = b2-4ac
Δ = -10502-4·6·1
Δ = 1102476
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:
$\sqrt{\Delta}=\sqrt{1102476}=\sqrt{4*275619}=\sqrt{4}*\sqrt{275619}=2\sqrt{275619}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1050)-2\sqrt{275619}}{2*6}=\frac{1050-2\sqrt{275619}}{12} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1050)+2\sqrt{275619}}{2*6}=\frac{1050+2\sqrt{275619}}{12} $

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