0.2(3x+1)0.1x=0.5(x+2)0.8

Solution for 0.2(3x+1)0.1x=0.5(x+2)0.8 equation:

0.2(3x+1)0.1x=0.5(x+2)0.8

We move all terms to the left:

0.2(3x+1)0.1x-(0.5(x+2)0.8)=0

We multiply parentheses

0x^2+0x-(0.5(x+2)0.8)=0


We calculate terms in parentheses: -(0.5(x+2)0.8), so:

0.5(x+2)0.8

We multiply parentheses

0.4x+0.8

Back to the equation:

-(0.4x+0.8)


We add all the numbers together, and all the variables

x^2+x-(0.4x+0.8)=0

We get rid of parentheses

x^2+x-0.4x-0.8=0

We add all the numbers together, and all the variables

x^2+0.6x-0.8=0

a = 1; b = 0.6; c = -0.8;
Δ = b2-4ac
Δ = 0.62-4·1·(-0.8)
Δ = 3.56
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.6)-\sqrt{3.56}}{2*1}=\frac{-0.6-\sqrt{3.56}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.6)+\sqrt{3.56}}{2*1}=\frac{-0.6+\sqrt{3.56}}{2}
$