2x+40=2/11(x)

Solution for 2x+40=2/11(x) equation:

2x+40=2/11(x)

We move all terms to the left:

2x+40-(2/11(x))=0


Domain of the equation: 11x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

2x-(+2/11x)+40=0

We get rid of parentheses

2x-2/11x+40=0

We multiply all the terms by the denominator


2x*11x+40*11x-2=0

Wy multiply elements

22x^2+440x-2=0

a = 22; b = 440; c = -2;
Δ = b2-4ac
Δ = 4402-4·22·(-2)
Δ = 193776
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:

$\sqrt{\Delta}=\sqrt{193776}=\sqrt{16*12111}=\sqrt{16}*\sqrt{12111}=4\sqrt{12111}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(440)-4\sqrt{12111}}{2*22}=\frac{-440-4\sqrt{12111}}{44}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(440)+4\sqrt{12111}}{2*22}=\frac{-440+4\sqrt{12111}}{44}
$