(5/10)x+2=3+2x/5

Solution for (5/10)x+2=3+2x/5 equation:

(5/10)x+2=3+2x/5

We move all terms to the left:

(5/10)x+2-(3+2x/5)=0


Domain of the equation: 10)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+5/10)x-(2x/5+3)+2=0

We multiply parentheses

5x^2-(2x/5+3)+2=0

We get rid of parentheses

5x^2-2x/5-3+2=0

We multiply all the terms by the denominator


5x^2*5-2x-3*5+2*5=0

We add all the numbers together, and all the variables

5x^2*5-2x-5=0

Wy multiply elements

25x^2-2x-5=0

a = 25; b = -2; c = -5;
Δ = b2-4ac
Δ = -22-4·25·(-5)
Δ = 504
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:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6\sqrt{14}}{2*25}=\frac{2-6\sqrt{14}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6\sqrt{14}}{2*25}=\frac{2+6\sqrt{14}}{50}
$