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(x-30)+40+(x-10)+1/3x=360
We move all terms to the left:
(x-30)+40+(x-10)+1/3x-(360)=0
Domain of the equation: 3x!=0We add all the numbers together, and all the variables
x!=0/3
x!=0
x∈R
(x-30)+(x-10)+1/3x-320=0
We get rid of parentheses
x+x+1/3x-30-10-320=0
We multiply all the terms by the denominator
x*3x+x*3x-30*3x-10*3x-320*3x+1=0
Wy multiply elements
3x^2+3x^2-90x-30x-960x+1=0
We add all the numbers together, and all the variables
6x^2-1080x+1=0
a = 6; b = -1080; c = +1;
Δ = b2-4ac
Δ = -10802-4·6·1
Δ = 1166376
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1166376}=\sqrt{4*291594}=\sqrt{4}*\sqrt{291594}=2\sqrt{291594}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1080)-2\sqrt{291594}}{2*6}=\frac{1080-2\sqrt{291594}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1080)+2\sqrt{291594}}{2*6}=\frac{1080+2\sqrt{291594}}{12} $
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