(3/4x+5)+(3x-7)+7x=180

Solution for (3/4x+5)+(3x-7)+7x=180 equation:

(3/4x+5)+(3x-7)+7x=180

We move all terms to the left:

(3/4x+5)+(3x-7)+7x-(180)=0


Domain of the equation: 4x+5)!=0

x∈R


We add all the numbers together, and all the variables

7x+(3/4x+5)+(3x-7)-180=0

We get rid of parentheses

7x+3/4x+3x+5-7-180=0

We multiply all the terms by the denominator


7x*4x+3x*4x+5*4x-7*4x-180*4x+3=0

Wy multiply elements

28x^2+12x^2+20x-28x-720x+3=0

We add all the numbers together, and all the variables

40x^2-728x+3=0

a = 40; b = -728; c = +3;
Δ = b2-4ac
Δ = -7282-4·40·3
Δ = 529504
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:

$\sqrt{\Delta}=\sqrt{529504}=\sqrt{16*33094}=\sqrt{16}*\sqrt{33094}=4\sqrt{33094}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-728)-4\sqrt{33094}}{2*40}=\frac{728-4\sqrt{33094}}{80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-728)+4\sqrt{33094}}{2*40}=\frac{728+4\sqrt{33094}}{80}
$