.05x+3=2/3x+1

Solution for .05x+3=2/3x+1 equation:

.05x+3=2/3x+1

We move all terms to the left:

.05x+3-(2/3x+1)=0


Domain of the equation: 3x+1)!=0

x∈R


We get rid of parentheses

.05x-2/3x-1+3=0

We multiply all the terms by the denominator


(.05x)*3x-1*3x+3*3x-2=0

We add all the numbers together, and all the variables

(+.05x)*3x-1*3x+3*3x-2=0

We multiply parentheses

3x^2-1*3x+3*3x-2=0

Wy multiply elements

3x^2-3x+9x-2=0

We add all the numbers together, and all the variables

3x^2+6x-2=0

a = 3; b = 6; c = -2;
Δ = b2-4ac
Δ = 62-4·3·(-2)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{15}}{2*3}=\frac{-6-2\sqrt{15}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{15}}{2*3}=\frac{-6+2\sqrt{15}}{6}
$