(1/6)x+x+2x=190

Solution for (1/6)x+x+2x=190 equation:

(1/6)x+x+2x=190

We move all terms to the left:

(1/6)x+x+2x-(190)=0


Domain of the equation: 6)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+1/6)x+x+2x-190=0

We add all the numbers together, and all the variables

3x+(+1/6)x-190=0

We multiply parentheses

x^2+3x-190=0

a = 1; b = 3; c = -190;
Δ = b2-4ac
Δ = 32-4·1·(-190)
Δ = 769
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{769}}{2*1}=\frac{-3-\sqrt{769}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{769}}{2*1}=\frac{-3+\sqrt{769}}{2}
$