(3/2)x+20+(3/2)x+10=180

Solution for (3/2)x+20+(3/2)x+10=180 equation:

(3/2)x+20+(3/2)x+10=180

We move all terms to the left:

(3/2)x+20+(3/2)x+10-(180)=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

(+3/2)x+(+3/2)x+20+10-180=0

We add all the numbers together, and all the variables

(+3/2)x+(+3/2)x-150=0

We multiply parentheses

3x^2+3x^2-150=0

We add all the numbers together, and all the variables

6x^2-150=0

a = 6; b = 0; c = -150;
Δ = b2-4ac
Δ = 02-4·6·(-150)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-60}{2*6}=\frac{-60}{12}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+60}{2*6}=\frac{60}{12}
=5
$