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

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

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

We move all terms to the left:

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


Domain of the equation: 2x+20)!=0

x∈R


We get rid of parentheses

2x+3/2x+20-180=0

We multiply all the terms by the denominator


2x*2x+20*2x-180*2x+3=0

Wy multiply elements

4x^2+40x-360x+3=0

We add all the numbers together, and all the variables

4x^2-320x+3=0

a = 4; b = -320; c = +3;
Δ = b2-4ac
Δ = -3202-4·4·3
Δ = 102352
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:

$\sqrt{\Delta}=\sqrt{102352}=\sqrt{16*6397}=\sqrt{16}*\sqrt{6397}=4\sqrt{6397}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-320)-4\sqrt{6397}}{2*4}=\frac{320-4\sqrt{6397}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-320)+4\sqrt{6397}}{2*4}=\frac{320+4\sqrt{6397}}{8}
$