3x+(1/2)x+x=120

Solution for 3x+(1/2)x+x=120 equation:

3x+(1/2)x+x=120

We move all terms to the left:

3x+(1/2)x+x-(120)=0


Domain of the equation: 2)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

3x+(+1/2)x+x-120=0

We add all the numbers together, and all the variables

4x+(+1/2)x-120=0

We multiply parentheses

x^2+4x-120=0

a = 1; b = 4; c = -120;
Δ = b2-4ac
Δ = 42-4·1·(-120)
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{31}}{2*1}=\frac{-4-4\sqrt{31}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{31}}{2*1}=\frac{-4+4\sqrt{31}}{2}
$