2x+(4/3)x+20=180

Solution for 2x+(4/3)x+20=180 equation:

2x+(4/3)x+20=180

We move all terms to the left:

2x+(4/3)x+20-(180)=0


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

2x+(+4/3)x+20-180=0

We add all the numbers together, and all the variables

2x+(+4/3)x-160=0

We multiply parentheses

4x^2+2x-160=0

a = 4; b = 2; c = -160;
Δ = b2-4ac
Δ = 22-4·4·(-160)
Δ = 2564
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:

$\sqrt{\Delta}=\sqrt{2564}=\sqrt{4*641}=\sqrt{4}*\sqrt{641}=2\sqrt{641}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{641}}{2*4}=\frac{-2-2\sqrt{641}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{641}}{2*4}=\frac{-2+2\sqrt{641}}{8}
$