40+1/3x+(x-10)+(x-20)=350

Solution for 40+1/3x+(x-10)+(x-20)=350 equation:

40+1/3x+(x-10)+(x-20)=350

We move all terms to the left:

40+1/3x+(x-10)+(x-20)-(350)=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+(x-10)+(x-20)-310=0

We get rid of parentheses

1/3x+x+x-10-20-310=0

We multiply all the terms by the denominator


x*3x+x*3x-10*3x-20*3x-310*3x+1=0

Wy multiply elements

3x^2+3x^2-30x-60x-930x+1=0

We add all the numbers together, and all the variables

6x^2-1020x+1=0

a = 6; b = -1020; c = +1;
Δ = b2-4ac
Δ = -10202-4·6·1
Δ = 1040376
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:

$\sqrt{\Delta}=\sqrt{1040376}=\sqrt{4*260094}=\sqrt{4}*\sqrt{260094}=2\sqrt{260094}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1020)-2\sqrt{260094}}{2*6}=\frac{1020-2\sqrt{260094}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1020)+2\sqrt{260094}}{2*6}=\frac{1020+2\sqrt{260094}}{12}
$