0.2(x+4)=2/10x+3

Solution for 0.2(x+4)=2/10x+3 equation:

0.2(x+4)=2/10x+3

We move all terms to the left:

0.2(x+4)-(2/10x+3)=0


Domain of the equation: 10x+3)!=0

x∈R


We multiply parentheses

0.2x-(2/10x+3)+0.8=0

We get rid of parentheses

0.2x-2/10x-3+0.8=0

We multiply all the terms by the denominator


(0.2x)*10x-3*10x+(0.8)*10x-2=0

We add all the numbers together, and all the variables

(+0.2x)*10x-3*10x+(0.8)*10x-2=0

We multiply parentheses

0x^2-3*10x+8x-2=0

Wy multiply elements

0x^2-30x+8x-2=0

We add all the numbers together, and all the variables

x^2-22x-2=0

a = 1; b = -22; c = -2;
Δ = b2-4ac
Δ = -222-4·1·(-2)
Δ = 492
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:

$\sqrt{\Delta}=\sqrt{492}=\sqrt{4*123}=\sqrt{4}*\sqrt{123}=2\sqrt{123}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{123}}{2*1}=\frac{22-2\sqrt{123}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{123}}{2*1}=\frac{22+2\sqrt{123}}{2}
$